Pre-Cal 40S (Winter '06)

The Adventure Continues ...

2006-06-30T15:51:00.000-05:00

Our adventures in blogging continue....

Watch for 3 new blogs going live September 6, 2006 ...

Pre-Cal 30S (Fall '06) (Grade 11)
Pre-Cal 40S (Fall '06) (Grade 12)
AP Calculus AB 2006-07 (Grade 12)

the last bob

2006-06-30T10:33:00.000-05:00

It has been a rollercoaster this semester from trig identities to logs. This math class was special and unique. t he only grade 12 precal class to blog =)... it sounds corny, but has made us all famous lol. good wishes to the grade twelves moving onto universit ;) . Hopefully everyone i knew i S3 for this precal class will take calculus.

HAVE AN AWESOME SUMMER GUYS AND GIRLS !!

i don't know why but i will always remember this math class just as a math class. Maybe because it represents that its the last math class for highschool. The learning experience was much more different compared to recent years. THIS math class was fu n and made you want to learn more =), i sure wanted to learn more.

So Long ...

2006-06-28T18:23:00.000-05:00

We had our graduation exercises today. A gentle push into the world for all of you. I hope you're leaving with the keys to your future in your hand.

I'm so glad we've had this time together,

Just to have a laugh or learn some math,

Seems we've just got started and before you know it,

Comes the time we have to say, "So Long!"

So long everybody! Watch this space in the fall for pointers to new blogs for each of my classes.

Farewell, Auf Wiedersehen, Adieu, and all those good bye things. ;-)

Student Survey Results

2006-06-18T23:00:00.000-05:00

We did a little survey in class the day we recorded our podcapsule. The results are below; 18 students participated. So, without any further ado, here are the results of our class' survey. Please share your thoughts by commenting (anonymously if you wish) below .....

Classroom Environment
The questions in this section were ranked using this 5 point scale:

Strongly Disagree	Disagree	Neutral	Agree	Strongly Agree
1	2	3	4	5

The bold numbers after each item are the average ratings given by the entire class.

1. The teacher was enthusiastic about teaching the course. 4.89

2. The teacher made students feel welcome in seeking help in/outside of class. 4.65

3. My interest in math has increased because of this course. 3.76
(This question answered by 17 students.)

4. Students were encouraged to ask questions and were given meaningful answers. 4.56

5. The teacher enhanced the class through the use of humour. 4.50

6. Course materials were well understood and explained clearly by the teacher. 4.22

7. Graded materials fairly represented student understanding and effort. 4.17

8. The teacher showed a genuine interest in individual students. 4.22

9. I have learned something that I consider valuable. 4.33

10. The teacher normally came to class well prepared. 4.83

Overall Impression of the Course
The questions in this section were ranked using this 5 point scale:

Very Poor	Poor	Average	Good	Very Good
1	2	3	4	5

1. Compared with other high school courses I have taken, I would say this course was: 4.33

2. Compared with other high school teachers I have had, I would say this teacher is: 4.78

3. As an overall rating, I would say this teacher is: 4.78

Course Characteristics
The questions in this section were ranked using this 5 point scale:

Very Easy	Easy	Average	Difficult	Very Difficult
1	2	3	4	5

1. Course difficulty, compared to other high school courses: 3.72

2. Course workload, compared to other high school courses: 3.67

3. Hours per week required outside of class:

0 to 2	2 to 3	3 to 5	5 to 7	over 7
17%	17%	22%	28%	17%

4. Expected grade in the course:

F	D	C	B	A
0%	29%	47%	18%	6%

Specific Feedback
What was your best learning experience in this course?

Learning to learn
Learning real life applications (2)
"Learning is a conversation."
Circular Functions
Combinatorics
Logs and Exponents
Transformations
Learning new formulas
Conic Sections

People helping each other
Math dictionary (4)
Scribes
Blogs (2)
Learning was fun and exciting
Probability
Group work (7)
Learning cool things

Stories
Wiki Assignment
None (2)
Everything
Fun classes
Learning grading methods
Pre-tests
Bubble tea
Help provided by teacher

What was your worst learning experience in this course?

Probability (2)
Conic Sections
Sequences
Identities
Learning that I didn't study for tests
Transformations

Logs and Exponents (2)
Quizzes
Online quizzes (5)
Finding links for del.icio.us box
Not understanding lesson (2)
BOB (Blogging On Blogging)

Long lectures
None (2)
Combinatorics (2)
Falling asleep in class
Bubble tea
Wiki Assignment

What changes would you suggest to improve the way this course is taught?

None (4)
More pre-test review
Be more approachable
Make a wiki notebook, edited yearly
Collect homework
Course was well taught (2)
More group work (3)
More bonus questions on tests

Give "Aim for I Rule! not I can get by" speech
"Person to person learning"
"Steady level of work"
Have 2 blogs; one for scribes, one for BOBs
Correct tests quicker
Go through every test
More assignments

Blog OK, but wiki and del.icio.us too much
Keep doing what you're doing
No online quizzes
Put del.icio.us button on blog
Homework only every second day

It's interesting to compare the items that were considered both the worst and best learning experiences. Also, take a look at the list of worst learning experiences compared to suggestions for next year. Help me do a better job next year by commenting on what you see here ....

Message in a Podcast: Our Podcapsule

2006-06-16T00:00:00.000-05:00

The students in our class wrote their final exam on Thursday June 15; they ended their high school mathematics careers. We made a podcast to celebrate! We left ourselves a "podcapsule" instead of a "time capsule." We left our future selves a message; hopefully it will help us improve our learning.

Here is our Podcapsule (8 minutes, 52 seconds). Please leave any questions, concerns, complaints, compliments, confusions, uncertainties, anxieties or other inquiries in the comments to this post.

You can also leave us an audio message if you wish. ;-) (You'll need a microphone.)

This isn't our last post yet -- I've got at least two more I want to get out before the end.

One of the LAST posts of this year

2006-06-15T08:20:00.000-05:00

Well its the morning of the exam, and its kind of late, but hope you all had a good nights rest and eat breakfast! Your mind needs the energy to think efficiently so good luck out there as we aim for the HIGH HIGH HIGH SUCCESS LEVEL! Remember what we said guys and girls? 80-100 no exception. Lets prove that we're capable to do what we were aiming for.

Scribe guY

2006-06-13T03:30:00.001-05:00

la la la

Scribe guY

2006-06-13T03:30:00.000-05:00

For today, I was the scribe. But I realized that I was the scribe at the end of second period class so I didnt' actually write anything down. So, I'll do my best to explain some of the stuff we covered for today's class. And sorry for the probably late post that you guys were expecting. But you should get used to that when it's my scribe.lol

Anyways. Back to the blogGing.

For second period, we corrected our tests
-Logarithms and Exponents
-Counting
-Probability

If I thoroughly explain everything, I probably won't sleep tonight. So sorry guys. I'm just going to give out the answers.

Part II

Multiple Choice

1) If $5000 is invested in an account that pays 6% compounded continuously, how long will it take the balance to grow to $75000?

Answer: B) 6.8 years

7500=5000(model)^0.06t

Ln(3/2)=0.06t

Work that out and you get 6.8 years.

2) The expression e^lnx-lny is equivalent to:

Answer:

D) x^2/y

3) In solving the expression 3^2x-1=4^x, which of the following is the exact solution?

Answer: C) log3/log9-log4

Turn it into log formation

2xlog3-log3=xlog4

2xlog3-xlog4=log3

X(2log3-log4)=log3

Log3/2log3-log4

Log3/log9-log4

4) The solution to the expression 2(3)^3=26 could be:

Answer: A)logbase3(13)

3^x=13

Logbase3(13)=x

5) The graph to the right is of f(x)=b^x+k

Answer: D) 0

(sorry guys. Don’t know how to explain this)

6) the graph at right represents the function y=logbase3(x). If the coordinates of C are (9,0), then the area of Triangle(ABC) is:

Answer: 8units^2

Since you know y=logbase3(x), you substitute 9 for x, since the x value is nine. Figuring the log out, it turns out t o be 2 after substitution. A is 1.

7) If logbase3(4)=X, then logbase3(64) equals:

Answer: D) 3x

8) A culture of bacteria doubles every 20 minutes. Which of the following is an expression representing the time t, in minutes, it takes for the original amount of bacteria in this culture to triple?

Answer: D) 3=2^t/20

A=Ao(m)t-p

Ao=Ao=2^t/20

3=2^t/20

Part III Short Answer

1) If log(x-2) + log(x+7)=1, then x must equal:

Log(x-2)(x+7)=1

X^2+5x-14=10

X^2+5x-24=0

(x+8)(x-3)=0

X= - 8 x=3-----answer

REJECT

Because if you plug it back in, it’ll turn out to be zero.

2) Correct to four decimal places, the value of logbase3(7/5) must be:

Log(7/5)/log3=0.03063

3^x=7/5

Xln3=ln(7/5)

X=ln(7/5)/ln3

=0.30627

3) Too lazy to draw the graph so I’ll just give you the answer. Sorry.lol

Answer:

2^x=y

2^2.36=c

5.1337=c

Part IV Problem Solving

1) Solve for x in each of the following. Write your answers accurate to 4 decimals places.

A)e^2x-1=6 b)logbase8(x-4)=1-logbase8(x+3)

Ln(e)^2x-1=ln6 logbase8(x-4)+logbase8(x+3)=1

2x-1=ln6 logbase8(x-4)(x+3)=1

2x=ln6-1 x^2-x-12=8

X=ln6-1/2 x^2-x-20=0

X=1.3959 (x+4)(x-5)=0

X=-4 x=5

REJECT

because plugging it back in will give 0.

2) a colony of what-cha-ma-call-its has a population of 8500. the population is increasing at a rate of 2.5% per year.

a)what will be the population of the colony in 15 years?

P=8500e^0.025t

P=8500e^0.025(15)

P=12367.4270

=12367 what-cha-ma-call-its

b) when was the population 7000?

7000=8500e0.025t

14/17=0.025t

Ln(14/17)=0.025t

(1/2)ln(14/17)=t

T= - 7.76624

- 7.7662 years ago

BONUS

Solve for x exactly without using a calculator:

3x^2e^( - x) = 12e^ -x

X^2=12e^-x/3e^-x

X^2=4

X^2-4=0

(x+2)(x-2)=0

X= +/- 2

AND THAT’S THE FIRST TEST DONE. That took about an hour and a half. Haha. Now, Just two more tests and two rehearsal exam things………

TEST ON COUNTING

Part II

1) A committee of 4 is to be selected from 4 boys and 3 girls. If both boys and girls must be on this committee, and that is the only restriction, the number of committees that could be formed is:

Answer: B)34

4c1 * 3c3+ 4c2*3c2+4c3*3c1

= 34

2)the number of arrangements of the word COMMUNICATION is:

Answer:C) 13P13/(2!)^5

2 c’s

2 o’s

2 m’s

2 n’s

2 I’s

3)all phone numbers in Brandon begin with either 948 or 912. the total possible phone numbers available for this community is given they all will be seven digits, with no restrictions on the last four digits is:

Answer: B)2*10^4

4)The fifth degree term in the expansion of (y-3)^7

Answer: A)189y^5

7c2 y z^5( - 3)^2

21 * y^5 * 9

189y^5

5)A class of 15 students are to be divided up so 5 students are in group 1, 3 studnets are in group 2, and 7 students are in group 3. the number of ways this can be done is:

Answer: 360 360

7) A juke box has a button for each of the 26 letters of the alphabet, and for each digit 1-9. in order to select a song, you must first enter two letters for the CD, and one digit for the track to be played. The number of songs this juke box could play is:

Answer: b)26*26*9

8)The middle term of the 7^th row of Pascal’s Triangle is given by:

Answer: c)6C3

Part III Short Answer

1)If n!/(n-2)!=420, the value of n, must be:

n(n-1)(n-2) / (n-2)=420

n(n-1)=420

n^2-n-420=0

(n-21)(n-+20)=0

n= 21 n= - 20

Rejected plugging it in gives 0.

2)in the expansion of (x+2/x)^6, the value of the constant term is:

P+q=6

p-q=0

2p=6

P=3

6c3 (x)^3 (y)^3

6! / 3!3! * 8

20*8

160

3)A car dealer is displaying 5 cars in his showroom. Each car is a different model. Three of the cars are teal in colour, on car is red, and the other is black. The 5 other cars are to be arranged in a line, with none of the teal cars directly beside each other. There are ? possible arrangements of these cars.

3 2 2 1 1 = 12

T T T

4)Four books, W, X, Y and Z are placed on a shelf. The number of possible arrangements if the books W and Z must stay together, but not necessarily in that order is?

3!2!= 12

PART IV PROBLEM SOLVING

(This is taking too long..sigh..)

1)Algebraically determine the value for n in the expression 7(n-1)! / (n-2)! = 6(n+1)! / n!

7(n-1) = 6(n+1)

7n-7=6n+6

N=13

2)A class of 18 students includes 8 boys and 10 girls.

a)if the girls line up for a picture, how many pictures are possible if Joan, Jennifer and Jillian, 3 of the girls in the class stay together as a group in the picture?

8!3!= 241 920

b) How many pictures are possible if Joan, Jennifer and Jillian are not together, as a group, in any of the photographs?

10!-8!3!= 3 386 880

c) The 18 students sit in a classroom with 20 identical desks. Express the number of possible arrangements of students as a factorial.

20! / 18!2! OR 20! / 2!

d) four students are going to be selected to represent the class in a school relay. However, the group of four must have more girls represented than boys. How many groups could be formed under these conditions?

10c3 * 8 c1 + 10c4

120 * 8 + 210

960 + 210

1170

3)a chorus consists of 6 singers in red and 6 singers in sliver costumes. In how many ways can they be arranged:

a)in two rows, the red behind the silver?

6!6!= 518 400

b)in a ring facing the centre, the colours alternating?

5!6!= 86 400

BONUS

The Chinese mathematician Shi Cheh called Pascal’s Triangle by another name; what was it?

THE PRECIOUS MIRROR OF THE 4 ELEMENTS

AND THAT’S THE SECOND TEST DONE. ONE MORE AND TWO EXAM REHERSEALS LEFT.HAHA.YESSS. SLEEP TIME IS SOON.

No body really cared about probability in class it seems because no one raised their hand up for help. And I’m weak in probability so don’t expect any good explanations. lol. Sorry.

PROBABILITY

PART II MULTIPLE CHOICE

1)a standard six-sided fair die is tossed twice. Find the probability of getting a 2, 4 or 6 on the first toss and a 2, 3, or 5 on the second toss.

Answer: a)1/4

2)A jar contains 5 red and 7 blue marbles. What is the probability of pulling out 2 blue marbles in a row, without replacement?

Answer: 0.318

3)A box of eight razor blades contains two defective blades. If two blades are drawn at random, with the first not replaced, what is the probability that exactly one of the two blades will be defective?

Answer: a) 3/7

4)one card is drawn is drawn at random from a deck of 52 cards. What is the probabitlity of drawing an ace or a diamond?

Answer: b) 4/13

4/52 + 13/52 – 1/52(ace of diamonds) = 4/13

5)A rack contains 15 dresses. Five of the dresses are blue, six are green, and 4 are yellow. If selling each of the dresses is equally likely, what is the probability that if six dresses are sold, exactly two will be green?

Answer: a) 6c2 * 9c4/15c6

6)In a car lot, 25% of the inventory are SUV’s, and 75% are passenger cars. 80% of the SUV’s, and 65% of the passenger cars, have air conditioning. What is the probability that a chosen vehicle will be an SUV given the vehicle has air conditioning?

Answer: 0.29

7) A combination lock requires the owner to choose three numbers from 1-40 in order to open the lock. If numbers can be repeated, the probability of an individual correctly guessing the combination to this lock is:

Answer: c) 1/40^3

8)Using the diagram to the right, there are 7!/4! * 3! Possible routs from A to C, given the shortest routes from A to C are traveled. What is the probability that a chosen route will pass through point B, in attempting to get from A to C?

Answer: c)4/7

PART III SHORT ANSWER

1)Events A and B have the following probabilities of occurring; 0.2=P(A), 0.5=P(B). if these events are mutually exclusive, the value of P(AorB), correct to the nearest tenth.

P(AorB)=P(A) + P(B)

0.2+0.5

0.7

2)Using the word, FOOD, the probability that an arrangement of this word will begin with the two O’s if all letters are used, correct to the nearest hundredth, must be:

2! * 2! / 2! ----ways the ‘words’ begin with 2 O’s

4! / 2!---arrange letters in food

=2/12

=1/6

=0.17

3)a box contains 100 computer floppy diskettes. The probability of a single diskette being faulty is 0.005. correct to the nearest hundredth, the probability that exactly two diskettes in the box will be faulty is:

100! / 98! 2! (0.005)^2 (0.995)^98

= 0.0757

4)Eight students of different heights are seated randomly around a circular table. The probability that the two tallest students are sitting next to each other is:

Arrange 8 students in a circle

(8-1)!

# of ways the 2 tallest are together

(7-1)!2!

P(2 tallest together) = 6!2!/7!

=2/7

0.2857

PART IV PROBLEM SOLVING

1)Rupert has either milk or cocoa to drink for breakfast with either oatmeal or pancakes. If he drinks milk, then the probability that he is having pancakes with the milk is 2/3. the probability that drinks cocoa is 1/5. if he drinks cocoa, the probability of him having pancakes is 6/7.

a) List a sample space of probabilities using a tree diagram or any other method of your choice

m-milk

c-cocoa

o-oatmeal

p-pancakes

P(mo)= 4/5 * 1/3 = 4/15

P(mp)= 4/5 * 2/3 = 8/15

P(co)= 1/5 * 1/7 = 1/35

P(cp)= 1/5 * 6/7 = 6/35

(draw the tree guys. I’m too lazy right now. You know how…)

b) find the probability that Rupert will have oatmeal with cocoa tomorrow morning.

P(co) = 1/5 * 1/7 = 1/35

2)Josh has purchased a male rabbit and a female rabbit. His research tells him that the breeding conditions the two rabbits will be exposed to dictate that the probability of a single offspring being male, M is 0.53, and female, F, is 0.47.

a) the first litter of rabbits produced 12 offspring, 4 of which were male. Correct to the nearest hundredth, what is the probability of this occurring?

12c4 (0.53)^4 (0.47)^8

= 0.0930

= 93%

b)in the next litter of rabbits, Josh makes the following probability calculation:

7c5(0.47)^5 (0.53)^2 + 7c6 (0.47)^6 (0.53) + 7c7(0.47)^7

This is the probability of having at least 5 females in a litter of 7 rabbits.

3)Five members of a mixed curling team including 2 females and 3 males. Only 4 can be chosen to play in a game.

a)what is the probability that all of the males will play the game?

3c3 * 2c1 / 5c4

=1*2/5

=2/5

b) if the five team members line up for a picture, what is the probability the two females will stand together?

4!2! / 5!

= 2/5

AND THAT’S THE THIRD TEST DONE. YES. SO CLOSE TO THE FINISH. YOU GUYS ARE PROBABLY NOT READING THIS BUT THAT’S OK! I wish I had a scanner. This would be so much easier.

EXAM REHERSEAL COUNTING

1)The number ofdifferent arrangements of 3 boys and 4 girls in a row, if the girls must stand together, is represented by:

A) 3!*4!

Because since you put the four girls in a bag, you have 3 boys plus the bag with gives you 4! Times the girls in the bag which is also 4!.

2)The students in a music department have practiced 6 contemporary and 5 traditional choruses. For their concert, they will choose a program in which they present 4 of the contemporary and 3 of the traditional choruses. How many different programs can be presented, if the order of the choruses does not matter?

c) 150

6c4 * 5c3 = 150

3)all telephone numbers are preceded by a 3-digit area code. In the original Bell Telephone System of assigning area codes, the first digit could be any number from 2 to 9, the second digit was either 0 or 1, and the third digit could be any number except 0. in this system, the number of different area codes possible was

8*2*9

=144 codes

4)a paperboy who delivers papers on his bike can travel only on the trails represented in the diagram. The number of different trails that the paperboy can take to get from house A to house B with out backtracking is:

*refer to pascal’s triangle.

5)a) how many groups of 3 chairs can be chosen from 7 chairs if the chairs are all different colours?

7c3=35

b)How many different ways can 7 chairs be arranged in a row if 2 of the chairs are blue, 3 are yellow, 1 is red, and 1 is green? (Assume that all of the chairs are identical except for colour.)

7!/2!3! = 420 ways

ONE MORE TO GO

……..

……

….

...

EXAM REHEARSAL EXPONENTS AND LOGARITHMS

1)The expression 3logbase(x) – 1/2logbase(y) is equivalent to:

b)log(x^3/ square root (Y)

2)If 27^x = 9^y-1, what is y in terms of x?

a)3/2+1

(3)^3X = 3^2Y-2

3x=2y-2

3x+2=2y

3x/2=y

2)f logbase(b)3 and logbaseb(5)=q, express logbase(b) cubed root of 0.6.

Logbase(b)(0.6)^1/3

1/3logbase(b)(3/5)

1/3logbase(b)3 * logbase(b)5

1/3(p-q)

1/3p – 1/3q

4)solve for x: logbase(3)x=2 – logbase(3)(x+2)

Logbase(3)x + logbase(3)(x+2) = 2

Logbase(3)(x)(x+2)=2

X^2+2x=2

X^2+2x=9----use quadratic formula

(not sure if this is right. Check with mr.k just in case if you don’t agree with it)

5)the Department of health in a developing country has issued a warning that the number, N, of reported cases of a certain disease is increasing expontentially at an annual rate of 9%. The first years that statistics were collected, 5000 cases were reported.

a)Calculate the approximate number cases reported after 7 years.

9388 cases (sorry guys. Don’t know how to get this but this is the answer)

b)how many years will it take for the number of reported cases to rise to 11 000?

8.7606 years(sorry guys. Don’t know how to get this either)

AND I’M FRIKKIN DONE. Sorry for the language. Anyways, this took too long as I’ve predicted and I did explain some stuff which I said I wouldn’t do but I knew I wouldn’t follow through on my words.haha.

ANYWAYS.

The next scribe will be.

Eehhhh I’ll pick tomorrow. guY =)

end of the course-afternoon

2006-06-11T20:47:00.000-05:00

So today there are two scribes. Marquin for the morning and me for the afternoon. This afternoon we FINALLY finished the course. This afternoon we wrapped up the last unit and also did a exam rehearsal for the Circular Functions unit .

So here's the notes...

Sigma Notation: A shorthand way to write a series.
example

is the capital sigma (from greek alphabet); means "sum"

- supbscript n=1 means "start with n=1 and evaluate 2n-3"

- superscript 4 means "keep evaluating (2n-3) for successive integral values of n, stop when n=4; then add all the terms."

- (2n-3) is the implicit definition of the sequence

Geometric Series: the sum, to the n^th term, of the terms in a geometric series given by:

- Sn is the sum of the first n terms in a geometric sequence

- n is the rank of the n^th term

- t₁ is the first term

- r is the common ratio

Infinite Geometric Series

- when >1, the infinite sum of a geometric series grows without bound _ it DIVERGES

- when <1,>CONVERGES

END OF COURSE

the next thing we did was the exam rehearsal for ciruclar functions.

1) Through how many radians does the minute hand of a clock turn in 24 minutes.

A) 0.2 pi

B) 0.4 pi

C) 0.6 pi

D) 0.8 pi

2) If cos theta = -3/4 and tan theta <0.>

A) -4/5

B) -root 7 / 4

C) 4 root 7 / 7

D) root 7 / 4

solution:

3) If y= 2 cos(1/2X) - 3, what is the minimum value of y?

-5

solution: since the graph lies on -3 and the amplitude is two the minimum value is -5. (-3) - 2 = -5

4) if theta=arcos(- root2 / 2), find the value(s) of tan theta?

tan= sin / cos

= -1

5) The tide at a boat dock can be modeled by the equation d(t) =-2cos(pi/6T)+8

a) How deep is the water at high tide? low tide?

since the amplitude is 2 you have to add 2 to 8 to get the high tide and subtract 2 from 8 to get the low tide.

low tide: 8-2=6

high tide: 8+2=10

b)How lond is it between one high tide and the next?

period=2pi/B

pi / 6 = 2 / period

12 = period

c) For how many hours, between t=0 and t=12 is the tide at least 7 feet deep?

7 = -2 cos pi/6T + 8

-1 = -2cos pi/6T

1/2=cos(pi/6T)

theta=pi/6T

1/2=cos theta

since cos 1/2 is equal to pi/3 and 5pi/3 on the unit circle, we plug it in for cos theta

theta = pi/3

pi/6T = pi/3

T=2

theta = 5pi/3

pi/6T = 5pi/3

T=10

now you find the difference of 10 and 2...

10-2=8 hrs

The tide will be at a dept of atleast 7 ft deep for 8 hrs.

That marks the end of my scribe. Don't mind if i say it again but i just love saying it.. THIS MARKS THE END OF THE COURSE.. haha.. anyways i don't know whole will be scribe tommorow since there was two scribes today. whose gonna pick? me or marquin? we'll just see tommorow..

Sequences and Series

2006-06-10T14:59:00.000-05:00

I guess blogger works right now and im hopping that im not going to lose this work. I should have known that saving thing. Anyways here is what we did Wed. Morning:
A superball is dropped from a height of 160 feet. It bounces to (3/5) of its height after it hits the ground. What is the total vertical distance it has traveled when:
a) it hits the ground for the 4th time?

As you can see that this question have 3 ways to answer.
b) It comes to a rest?

And there is 3 ways to answer this question too.
Soo is the infinite sum of the series
t1 is the first term
r is the common ratio.
And that was all we did in the morning and this what we did in the afternoon:

And that was all we did in the afternoon. Oh we did notes on the morning too. Here they are:
Common Ratio (r): I) the number that is repeatedly multiplied to successive terms in a geometric sequence.
ii) from the implicit definition (r) is the base of the exponential function.
To find the nth term in a geometric sequence
tn=t1r^(n-1)
tn is the nth term
n is the rank of the nth term
t1 is the first term
r is the common ratio
Sigma notation: A shorthand way to write a series.
Oh and there is one thing before I finish. The smallest number a calculator can get to is 9.999999999999999*10^(-99)
And the largest number a calculator can get to is 9.999999999999999*10^(99).
Example:
(3/5)^4=.1296
(3/5)^41=8.02*10^(-10)
(3/5)^100=6.53*10^(-23)
(3/5)^1000=0 (its not really 0 but the calculator shows 0 because this number is really small, smaller than that number that's why the calculator shows 0.)
Oh and Thales thought magnets had souls.
anyways that is all for my scribe, you guys have fun.

Create a Scribers Guide to Scribing

2006-06-08T10:15:00.001-05:00

Two days ago I received this email from a teacher friend of mine, Mr. Harbeck.

I would like to congratulate you[r classes]. Every scribe post I read is like taking part in your class. The students take tremendous pride in creating their posts and are all scribing at Hall of Fame levels. It would be interesting for them to reflect on their scribing and come up with a criteria for making the Hall of Fame. They know how much effort they put into their posts and what is worthy or not. As an outsider teaching 13 year olds who have no choice but to be in my class I do not see the devotion these scribes put forth every day. Everyone takes a turn and does a great job.

I would be curious if you did a "bob" on reflecting about scribe posts. The first to the last. What made a difference from the beginning of the course to the end. My topic was a dry etc one...this is how I spiced it up.

Could they create a scribers guide to scribing?

This is a talented bunch of students. Use them while you still have them. The Next bunch could be even better.

So here it is folks. You've more or less created this art form, now you can write the book on it. In the comments to this post answer these questions:

How do you go about writing a scribe post? Do you do anything differently in class when it is your turn to scribe? If so, can you describe what you do differently in class when you are scribe?

What makes a scribe post worthy of entry into The Scribe Post Hall Of Fame? Specifically, what should be included in the post for it to achieve this recognition?

Compare the first scribe post you wrote to the most recent one. What, if anything, did you do differently?

Your replies, and those of my other classes, will be collated and posted on a special page in The Scribe Post Hall Of Fame. Do yourselves proud. The world is watching; teach them how it's done. ;-)

Scribe are ME!

2006-06-06T22:20:00.000-05:00

hi, im the scriber for the day and ill start it off with a word problem on the board.

SCRIBE: Sequences & Series

2006-06-05T16:25:00.000-05:00

Well Michael ended off his scribe with a coded sequence to reveal who the next scribe with so I'll show you how we found out it was me. 14, 34, 39, 43, 47, 56, 61.

The differences between each numbers are as follows

20, 5, 4, 4, 9, 5

Since he told us that the differences represented the order of the alphabets, we get this word.

T E D D I E

So today's scribe is me Teddie!

Okay enough with the fun and game and lets get down to the stuff we did today cuz it was a whole lot.

THE MATH DICTIONARY

SEQUENCES & SERIES

Sequence: An ordered list of numbers that follows a certain rule or pattern.

Series: The sum of numbers in a sequence to a particular term.

Example: Given the sequence
3, 6, 12, 24, 48, . . .

S₁ = 3
S₂ = 3 + 6 = 9
S₃ = 3 + 6 + 12 = 21
S₄ = 3 + 6 + 12 + 24 = S₅ = 3 + 6 + 12 + 24 + 48 = S_n = 3 + 6 + 12 + 24 + 48 + . . . . t_n
S₅ Denotes the sum of the first 5 terms.
S_n denotes the sum of the first n terms.

ARITHMETIC SEQUENCE

I ) Recursive Definition: an ordered list of numbers generated by continually adding a value (the common difference) to a given first term.

II ) Implicit Definition: an ordered list of numbers where each number in the list is generated by a linear equation.

III ) Common difference (d):
a) the number that is repeatedly added to successive terms in an arithmetic sequence.
b) from the implicit definition d is the slope of the linear equation.

To Find a Common Difference

d = t_n -t_(n-1)

d is the common difference
t_n is any term in the sequence other than the first term.
t₁ is the first term in the sequence

Finding the n^the Term in an Arithmetic Sequence

t_n = t₁ + (n - 1)d

t_n is the n^the term in the sequence
t₁ is the first term n is the rank of the term
d is the common difference

Example: find the 51^>st term of the arithmetic sequence 11, 5, -1, -7, . . . .

Solution:

n = 51
t₁ = 11
d = -6

t₅₁ = t₁ + (n - 1)d
= 11 + (51 - 1)(-6)
= 11 + (50)(-6)
= 11 - 300
= -289

ARITHMETIC MEANS

The term(s) that falls between any two non-consecutive terms in an arithmetic sequence

Examples: Find the arithmetic means in each sequence

a) 1, _ , 25
there are 2 gaps for differences to occur
25 - 1 = 24
2d = 24
d = 12

1, 13 , 25

b) 3, _ , _ , _ , 1
there are 4 gaps for differences to occur
1 - 3 = -2
4d = -2
d = -0.5

3, 2.5, 2, 1.5, 1

Arithmetic Series: the sum of numbers in an arithmetic sequence given by:

S_n = (n/2) [ 2t₁ + (n - 1) d ]

d is the common difference
S_n is the sum to the n^the term
t₁ is the first term

Example:
find the sum to the 21^st term of the sequence
5, 9, 13, 17, . . .

Solution:
t₁ = 5
d = 4
n = 21

S_n = (n/2) [ 2t₁ + (n - 1) d ]
S₂₁ = (21/2) [ 2(5) + (21 - 1)(4) ]
S₂₁ = (21/2) [ 10 + 80 ]
S₂₁ = (21/2) [ 90 ]
S₂₁ = 21 [ 45 ]
S₂₁ = 945

GEOMETRIC SEQUENCES:

I ) Recursive: an ordered list of numbers generated by repeatedly multiplying a given first term by a particular value (the common ratio).

II ) Implicit: the ordered list of numbers where each number in the list is generated by an exponential function

And that is all for the notes in our dictionary, pew.

Starting off the afternoon we started off with some problems.

1. Find S_n for each arithmetic sequence

a) t₁ = 8
d = 5
S₂₅ = ?

S₂₅ = (25/2) [ 2(8) + (25 - 1) 5 ]
S₂₅ = (25/2) [ 16 + 120 ]
S₂₅ = (25/2) [ 136 ]
S₂₅ = (25) [68]
S₂₅ = 1700

b) t₅ = 15
t₁₉ = 43
S₁₀₀ = ?
d = ?
t₁ = ?

19 - 5 = 14 common differences required
43 - 15 = 28
14d = 28
d = 2

t₅ = t₁ + (5 - 1) d
15 = t₁ + (4)(2)
15 = t₁ + 8
7 = t₁

S₁₀₀ = (100/2) [ 2(7) + (100 - 1)(2) ]
S₁₀₀ = 50 [ 14 + 198 ]
S₁₀₀ = = 50 [212]
S₁₀₀ = 10600

2. Given S₁₂ = 360, t₁ = -3 find d
S₁₂ = (12/2) [ 2 t₁ + (12 - 1) d ]
360 = 6 [ 2(-3) + 11d ]
360 = -36 + 66d
396 = 66d
d = 6

3. Given S₃₀ = 480 t₁ = 4 find t₃₀

S₃₀ = (30/2) [ 2t_{1 + (30 - 1) d ]} 480 = 15 [ 2(4) + 29d ]
480 = 120 + 435d
360 = 435d
24/29 = d

t₃₀ = t₁ + (30 - 1) d
t₃₀ = 4 + 29(24/29)
t₃₀ = 4 + 24
t₃₀ = 28

4. x = 1, (1/2)x + 4, 1 - 2x form and arithmetic sequence what are the first 3 terms?

[(1/2)x + 4] - (x - 1) = (1 - 2x) - [(1/2)x + 4]
1/2)x + 4 - x - 1 = 1 - 2x - (1/2)x + 4
2x = -8
x = -4
x - 1 = -4 -1 = 5
(1/2)x + 4 = (1/2)(-4) + 4 = 2 1 - 2x = 1 - 2(-4) = 9

Finally I am done yay! And tomorrow's scribe is CALVIN!

SEQUENCES part 2!!!

2006-06-04T19:42:00.000-05:00

I would like to thanks MANNY to do the last scribe because I had a family problem. We started the first period class by doing problems on sequences. The following problems are as follows:In the afternoon class, we went over our online quiz. If you want to check your answers, you can see them at quizstar.com After that we went over more examples on logarithms, which took the rest of the class. This is my SCRIBE POST and I hoped everyone liked it. Since we are in the unit sequences, i'll use sequences to figure out tomorrows scribe. The difference will produce numbers and each number represents a letter in the alphabet. (eg. A = 1 , Z = 26). The sequence is 14, 34, 39, 43, 47, 56, 61. Have FUN!!!!!!!!!!

blogs on blogs

2006-06-02T10:55:00.000-05:00

So test today and I am still not comfortable with logs and exponents. I should have done all my homework! I am the same with regine, i forget that a log is an exponent and the whole model I dont even remember the formula. Well good luck guys.

loooogs

2006-06-02T01:29:00.000-05:00

i always forget. a logarithm is an exponent. No matter how much that's repeated, i still forget it. and that really screws me up(oops is that ok to say?) the word problems are also a killer especially the model one. where you don't have all the information. this unit reminds me of identities because you can get creative of how to solve a logarithm....and I didn't like identities either..

logs and exponents

2006-06-01T22:28:00.000-05:00

I understand this unit no problem so I like it. However once again I'm having trouble with the word problems. The exponential modeling types of question specifically frustrate me. I take way too long to do those questions. The ones I'm confident at doing are the laws of logarithm questions because I fully understand those. Another thing, I hope the test doesn't have any proof questions because I'm not comfortable with those questions either.

BOB

2006-06-01T21:08:00.000-05:00

i thing this unit was alright but graphs is the only problem im havin. newayz i think this will not be a problem after 2night because im gonna be studyin so hard and that means all u guys should do the same.

good luck on the test tomorrow but u r not goin to need it if u guys study hard and kno wut u guys r doin tomorrow.

B.O.B

2006-06-01T17:30:00.000-05:00

I'm finding this unit on neutral level for me. What I mean is that some parts and hard and the others are easy. The thing that I'm not sure about is when a graph is involved. I dont know how to look at these types of quesitons. Another thing that was bugging me was solving for x. We went over this in class today and I still have to look at it a little more to understand it. Overall Im okay with a really hard unit. As i always say, I have to look at thing differently and broader and ALSO watch out for those small errors taht i make.

Geometric Sequences / First Day

2006-05-31T20:41:00.000-05:00

Hey guys, it's Manny doing the scribe for today, cause yeah Michael couldn't so I volounteered for it. Today, we started our new unit Geometric Sequences, and thankfully, or sadly for some, it's our last. It was a lot to absorb without taking some form of notes, so I'll try my best to describe it to you guys.

So what is a sequence? A sequence is a list of numbers that follows a certain rule. That's what it is.

In the beginning of class, Mr.K put up a question with 6 parts that looked like this:

Find the next Four Terms in Each Sequence:

a) 4, 7, 10, 13, __, __, __, __

b) 3, 6, 12, 24, __, __, __, __

c) 1, 2, 4,__, __, __, __

d) 2, 5, 10, 17, 26, __, __, __, __

e) 2, 1, 3, 4, 7, 11 __, __, __, __

f) 77, 49, 36, 18, __, __, __, __

(We talked a lot about the first two.)

Well, at first we thought it was real easy, something done in like grade five. The first one (part a) we said, add 3 to the last number, easy as cutting butter. Well it was, but why is it 3? We all went speechless. Then we said that because taking two that are beside eachother and finding the difference between them will give us 3, so therefore it always has to go up by 3. So this is what we did.

7 - 4 = 3
10 - 7 = 3
13 - 10 = 3
And it just so happens that the 3 is called the common difference.

So this is what we have so far for part a) 4, 7, 10, 13, 16, 19, 22, 25.

Then, Mr. K, asked so what would the 11^th number be? We answered 34. It was right, and we did this by adding 3, to the last number three more times, since there are eight terms already.

It turns out Mr.K made a "mistake," he meant to ask for the 1001^th number. Well the 11^th, and 1001^th, or n in this case, are called ranks. One of us saw that it was one number less than the rank, multiplied by 3 and added to the first terms which was 4. The term is the output of the rank.

We came up with the formula t_n = t₁ + d(n-1).
Where t_n is the term of the number (or rank number),and Where t₁ is the first term, and
Where d is the common difference.
So therefore, t_n = 4 + 3(n-1).
And for the problem for the 1001^th looks like this t₁₀₀₁ = 4 + 3(1001-1), which equals out to a huge number bigger than googal =] (There is not as much atoms in the universe than the number googal 10¹⁰⁰).
This type of equation has a name called the Recursive Definition, because recursive means to apply some feature over and over again, here we are adding 3 all the time.

Well now that we know all this information, what is the initial term, Mr. K asks? Well since we added 3 to the last number to get the next, we can just subtract 3 from the first. And at the 0th rank, we have 1. This is what it looks like.

rank (n)	0	1	2	3	4	5	6	7	8...
term (t)	1	4	7	10	13	16	19	22	25...

Since we now know the 0th term, we can now write another equation. And come up with t_n = dn + t_o, therefore being t_n = 3n + 1. This equation is called the Impicit Definition, because implicit means that is implied, not obvious, and is in the subtext kind of like when some of us nodd or a raise an eyebrow to say hello to a friend while an adult would not understand it.

Then we graphed it to look like this.

We did this to show that from this pattern, given only 4 pieces of info, we can show it
symbollically - written as an equation
numerically - table of values
graphically -drawn out as a graph
Which is shown above. These are the 3 perspectives you can look at just about anything given at least one of them.

We are only learning about two types of sequences in this course. Arithmetic, and geometric. Arithmetic deals in adding/subtracting types of patterns, which was what we just did. Geometric is dealing with multiplication/division types of patterns.

Part b) 3, 6, 12, 24, __, __, __, __, is geometric.

- Will explain soon -

If anyone is curious the answers to the question above are:

a) 4, 7, 10, 13, 16, 19, 22, 25 [common difference 3]

b) 3, 6, 12, 24, 48, 96, 192, 384 [common ratio 2]

c) 1, 2, 4, 8, 16, 32, 64 [powers of 2 raised to the exponent of the (rank - 1)]

d) 2, 5, 10, 17, 26, 37, 50, 55, 72 [square the rank + 1]

e) 2, 1, 3, 4, 7, 11, 18, 29, 37 [sum of two side by side numbers, or fibonacci sequence]

f) 77, 49, 36, 18, 8, 8, 8, 8 [multiply digits]
or77, 49, 36, 18, 08, 0, 0, 0 [multiply digits, but have to have 2 digits]

We only did the first two today, and will not be looking at the last three parts, as they are different from arithmetic and geometric sequences.

Well, I have had a long day today. I was asked to volounteer last minute, and from the kindness of my heart I couldn't refuse, just like how I volounteered for this scribe today. But I am tired, and starving, I have had nothing but junk food since the past 14 hours, and it's way past my bedtime. I will continue the post tomorrow morning, and if not by the end of tomorrow. I'm sorry if this causes an inconvience to anybody and will be happy to fill you in if you want face to face.

The scribe for tomorrow's class will I guess be Michael since he didn't do it today, and if that's not possible, it'll be Master Teddie ;).

blogging for the test!

2006-05-30T22:59:00.000-05:00

Overall this unit wasn't all that bad. I guess the thing that confuses me the most is questions dealing with the base law, and comparing it to another base. Also what I have a little trouble on is gathering all the information from a long answer question, to help solve the problem. Well I guess that's about it, goodluck everyone!

blogpost

2006-05-30T22:33:00.000-05:00

the unit exponents and logarithms is less confusing to me than the last unit which is probability. i like this unit except on one part. the logarithm expression raised on an exponential expression often made me confused especially when a logarithm expression is raised on another logarithm expression. i get it after mr. k explains it to us but i forgot how it happen again after a while. well, that's my only big problem about this unit.

blogging on blogging

2006-05-30T22:30:00.000-05:00

Well this unit is going well for me. No bumpy roads except for the word problems, namely the ones with the least amount of information. Everything else I could do nicely. I don't have anything else to say except for the need of more word problems for practice

Rap-up of Exponents and Logarithm

2006-05-30T16:20:00.000-05:00

My last scribe ever...YAY! I'm pretty happy about that but I'm also disappointed that my own right hand man *ahem Jefferson betrayed me! So much for taking over the blog. Anyways I guess I'm on my own now and I should get right to it since we did so much work today, so here it goes...

In the morning class we got right down to business. Mr. K put up old exam questions on the board. There were two parts, a calculator and a non calculator part. The first part you were allowed to use a calcualtor:

1) 3^x+4 = 7^2x+1
2) If log_a2=0.3562 and log_a5=0.8271 show that log_a40=1.8957
3) A new car costs $24 000. It's value after t years is given by V=24000(0.8)^t
a.) Determine the value after 8 years.
b.) How many years will it take for its value to decrease to one-eighth of its initial value?

For the rest of the questions calculators weren't allowed:

4) A population triples every 7 days. Determine an expression (formula) for the total population after t days if the initial population is 800.
5) Evalute: log(100 sin(π/2))
6) State the range of f(x)=2^-x
7) Given:

1/2 ≤ log x ≤ 2
find the minimal and maximum value of x.

Solutions:

1)      log3^x+4 = log7^2x+1          ← change it as a log
     (x+4)log3 = (2x+1)log7        ← by the rules of logarithm
   xlog3+4log3 = 2xlog7+log7       ← distribute the brackets because the brackets are #'s
  xlog3-2xlog7 = log7-4log3       ← get all the #'s and x's together
x(log3-2xlog7) = log7-4log3       ← factor out the x
             x = (log7-4log3)      ← solve for x using a calculator
                 (log3-2log7)
             x = 0.8766

2) log_a40 = log_a(2³∙5)
          = log_a2^{3+log_a5
             = 3log_a2+log_a5
             = 3(0.3562)+(0.8271)
             = 1.0686+0.8271
             = 1.8957}


3.a) V(8)=24000(0.8)⁸
         = 4026.5318
3.b) (1/8)24000 = 3000
original value=24000 final=3000 therefore,

  3000 = 24000(0.8)^t
 24000     24000
 (1/8) = (0.8)^t
ln(1/8)=tln(0.8)
ln(1/8)=t
ln(0.8)
9.3189 ≈ t   ← using a calculator evaluate t


4) P =7days           A=A_°M^t/p
   M =3               A=800(3)^t/7 ← just substitute the values
  A_° =800
   t =?

5) log(100 sin(π/2))
   =log(100∙1)         ← since sin(π/2)=1
   =log(100)
   =2

6) f(x)=2^-x
So to solve this question we must think back on transformations.
We know that this is an exponential decay because the basic function (2^x) is an
exponential growth. Now in transformations when x is negative that means it "flips
over the y-axis" in other words the x-coordinates switch.
Therefore the answer must be (0,∞) because it only approaches 0 but never touches it.

7) 10^1/2 ≤ 10^logx ≤ 10²       ← changing it to base 10 will simplify it
    √10  ≤   x   ≤  100


Everyone was kind of stuck on that last question so Mr. K
tried to clear it up by giving us another question which
he thought was mildy amusing.
 log(logx)=2
       10²=logx
       100=logx
      10¹⁰⁰=x        which we know as Googol!
                    humorous? i didnt think so.

Now that took the whole morning class and in the afternoon class we wrote in our dictionary for the whole period. There were five pages of notes so for those students that missed class or didn't finish copying here it is:

PROOFS OF THE LAWS OF LOGARITHMS

Proof of the the product laws
Prove log_aMN = log_aM + log_aN

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬
   x=log_aM  IFF  a^x=M    
   y=log_aN  IFF  a^y=N      
▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬
log_aMN=log_a(a^x∙a^y)
      =log_aa^(x+y)
      =x+y
      =log_aM+log_aN
      Q.E.D

Proof of the quotient law
Prove log_a(M/N)=log_aM-log_aN

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬
   x=log_aM  IFF  a^x=M    
   y=log_aN  IFF  a^y=N      
▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬
log_a(M/N)=log_a(a^x/a^y)
      =log_aa^(x-y)
      =x-y
      =log_aM-log_aN
      Q.E.D

Proof of the power law
Prove log_aM^c=clog_aM

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬
   log_aM=b  IFF  a^b=M      
▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬
log_aM^c=log_a(a^b)^c
      =log_aa^cb
      =cb
      =clog_aM
      Q.E.D

Proof of the base law
Prove:

log_aM^c= log_bM
        log_ba
▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬
   log_aM=c  IFF  a^c=M      
▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬
    a^c=m
log_ba^c=log_bM
clog_ba= log_bM
        log_ba
     c=log_bM
        log_ba
 log_aM= log_bM
        log_ba
      Q.E.D

EXPONENTIAL MODELING

the basic function:

How we model a real life situation depends on how much information we are given.

case 1: Given a minimal amount of information (A, A_°, ▲t) we will create a model in base 10 or base e. (base e is prefered)
A is the amount at the end of the time period.
A_° is the original amount of the "substance".
MODEL is an exponential function that describes the growth (or decay) of the "substance".
t is the time required for the amount of the "substance" to change from A_° to A.

NOTE: Some "substances" are populations. The model is usually an exponential function of the form 10^k, or e^r where k and r are constants.

EXAMPLE: The population of earth was 5 billion in 1990. In 2003 the population grew to 6.3
a) Model the population as an exponential function.
b) What was the population of earth in 1967?

a)                let t=o in 1990.
A_°=5                                  A=A_°(model)^t
 A=603 (in 2003)                    6.3=5(model)¹³
▲t=2003-1990                       6.3 =(model)¹³
  =13                               5

From here you can go on in two different ways. It wouldn't really matter which way you choose because it will end up having the same answer.

if you choose base 10                            if you choose base e
    log(6.3/5)=log(model)¹³                      ln(6.3/5)=ln(model)¹³
    log(6.3/5)=13log(model)                      ln(6.3/5)=13ln(model)
1/13log(6.3/5)=log(model)                    1/13ln(6.3/5)=ln(model)
        0.0077=log(model)                           0.0178=ln(model)
       10^0.0077= model                               e^0.0178= model
therefore    P=5(10^0.0077t)                               P=5e^0.0178t

b)    If t=0 in 1990 then t=-23 in 1967.
if base 10                              if base e
P=5(10^0.0077(-23)                         P=5e^0.0178(-23)
P=3.3219                                P=3.3219

case 2: Given lots of information (A, A_°, M, p)A is the amount at the end of the time period t.
A_° is the original amount.
M is the "multiplication factor" or growth rate.
t is the time has passed.
p is the time period; time required to multiply by M once.

EXAMPLE 1: A colony of bacteria double every 6 days. If there were 3000 bacteria at the start of an experiment how many bacteria will there be in 15 days?

 A=?             A=A_°M^t/p
A_°=3000           =3000(2)^15/6
 M=2              =16970.5628
 t=15             ≈16971 bacteria
 p=6

EXAMPLE 2: The half life of carbon 14 is 5700 years. How much of a 10mg sample will remain after 4500 years?

 A=?             A=A_°M^t/p
A_°=10             =10(1/2)^4500/5700
 M=1/2            =5.8022 mg
 t=4500             
 p=5700

And that's that!! Phew...~~three~~ five small coffee cups later and I'M FINALLY DONE! YAY! Okay so now that I'm all finished and I'm in a cranky mood I will pick none other than michael as the scribe for tomorrow. Nah, enjoy!

Logs and Exponents Online Quiz

2006-05-30T15:26:00.000-05:00

The quiz goes online at 3:30 this afternoon. The deadline is 9:00 am Thursday morning. The unit test will be on Friday.

You can write the quiz here. You have 45 minutes to complete it. If you haven't yet registered at QuizStar then when you get to the site follow these instructions:

Click on the big yellow "Sign Up" arrow.

Use only your first name and last initial as indicated.

Pick a username that will allow me to easily identify you, i.e. first name and last initial.

Make up any password you like.

Click on [Register] then [Search] by teacher's name (kuropatwa) and you'll find me.

Click on the box next to Pre-Cal 40S and then [Register].

Follow the instructions on the screen.

Actually, if you read each page carefully, you'll see that the sign up process is very straight forward and self explanatory. If you hit any snags email me and we'll sort it out together.

Remember, the quiz is timed. You'll only have 45 minutes to complete it once you've begun. It consists of 11 multiple choice questions (1 mark each). DON'T PANIC. Take your time. I know you'll all do well. ;-)

Do Your Best!

Sisyphus' Bloggon the Blog 7

2006-05-30T08:17:00.000-05:00

Logarithms and exponents. A logarithm is an exponent. This unit isn't too bad, well not as bad as the time during the unit counting. I understand most of the simple short answer questions. The thing that I have trouble with, as do probably most other classmates, are the long answers. Yesterday's class doing the group work, was tough! We did not get a single question right. We just didn't understand the question. Fortunately a few groups were lucky to fluke the answer, but does not make sense to the problem. I think you should drop that assignment Mr.K ;). So, like the other units... It's those darn long answers!

Jefferson's FINAL SCRIBE - ACTION ON EXPONENTS AND LOGARITHMS!!

2006-05-29T21:11:00.000-05:00

HEY HO! Its scribe specialist and former scribe emperor, Jefferson presenting today's scribe post! Attention! The logarithm test will be on Friday, the wiki project will also be due on Friday.

Mr.K started off today's lesson with a quiz. For those who were absent here is a copy of the quiz:

Now lets go through the quiz together shall we?

(1) Find log₁₀(0.001)

solution:

log10(0.001) = -3

(2) Find log₂(64) + log₃(9)

solution:

log2(64) + log3(9) = 6 + 2
log2(64) + log3(9) = 8

(3) 3^x = 7; which is true?

(a) 7= log_x(3) (b) x = log₇(3) (c) 3 = log_x(7) (d) 7 = log₃(x) (e) x=log₃(7)

solution:
REMEMBER a log is an exponent. "7" is the power so the log format would be:
log_basepower = exponent
so the answer would be (e) x=log₃(7)

(4) if log_a(25) = 4, what is a?

(a) 1/5 (b) sqr(5) (c) 5 (d) log2 (5) (e) none of these

Log_a(25) = 4
a = sqr(5)

(5) find log₁₀₀(1 000 000)

log100(1000000) = 3

(6) If log₁₀ 2 ≈ 0.301, which of these is the closest to the log₁₀(2000)?

(a) 0.6 (b) 3 (c ) 6 (d) 30 (e) 60 (f)6

log₁₀(2000) = log₁₀(2 x 1000)
log₁₀(2000) = log10(2) + log₁₀ (1000)
log₁₀(2000) = 0.301 + 3
log₁₀(2000) = 3.301

so the answer is (b) 3

(7) If log₁₀ 2 ≈ 0.301, which of these is the closest to log₁₀(8)?

(a) 0.3 (b) 0.6 (c )0.9 (d) 1.2 (e) 2.4 (f) 6

log₁₀8 = log₁₀(2)³
log₁₀8 = log₁₀(2)³
log₁₀8 = 3log₁₀(2)
log₁₀8 = 3(0.301)
log₁₀8 = 0.903

and the answer is (c) 0.9

(8) If log₁₀ 2 ≈ 0.301, which of these is the closest to log₂(10)?

(a) 0.5 (b) 1 (c ) 3 (d)5 (e)20 (f)50

log2(10) = log2(2*5)
log2(10) = log2(2) + log2(5)
log2(10) = 1 + 2.32
log2(10) = 3.32

and the answer is (c) 3

(9) find 2^log₂(17)

solution:

17

(10) find 4^log₂(3)

4^log₂(3) = 9
(2²)^log₂(3) = 9
2^2log₂(3) = 9
Log₂(3)² = 9
2log₂(9) = 9

*not sure if this is correct Mr k , or anyone else please correct help me correct this*

(11) if log₃((10) = k, then log₉(10) =

(a) 2K (b) k/3 (c )k/2 (d)k² (e)2.4 (f)sqr(K)

log₃(10) = K
3^K=10

*since we don't know what the log9(10) is we make it equal to any variable we name, lets call it "n" courtesy of Mr k.*

log₉(10) = n
9ⁿ = 10

3^K = 9ⁿ
3^K = 3²ⁿ
K = 2n
k/2 = n

and the answer is k/2

After the quiz, we were put into groups and faced a difficult word problem. Again, for those who were absent, here is a copy of the hand out:

AND NOW LETS GO OVER THE QUESTIONS!!

(a) The intensity of thunder is 10¹² times the intensity of barely audible sound. What is the decibal level of thunder?

I = 10¹² * 10^-12
I = 1

* I = intensity of thunder * barely audible sound *

D = 10log(1/10^-12)
D = 10log10¹²
D = 120

(b) If the decibal levels of a subway train entering a station and normal conversation are 100 dB and 60 dB, respectively, how many times as intense as normal conversation is the noise of the subway train?

D_t = 100
100 = 10log(I * 10¹²)
10 = logI + log10¹²
-12 + 10 = logI + logI
-2 = logI
10^-2 = I_t

D_c = 60
60 = 10log(10¹² * I)
6 = log(10)¹² + logI
6 = 12 + logI
-12 + 6 = logI
-6 = logI
10^-6 = log I_n

I_t/I_n = 10^-12/10^-6
= 10⁴
=10000 times more intense

*I had just resently obtain the last bit of the answers after class, unfortunately I do not understand what's going on. Hopefully Mr K will explain this solution when he sees this scribe*

WELL I AM DONE!! Took me way too long to do this scribe. Operation rebellion will commence by a new member of the anti scribe take over movement, namely me =). My first victim will be NONE OTHER THAN JANET!!
that is all, *runs away*

later days, Jefferson

3D Tic-Tac-Sunday

2006-05-28T22:14:00.000-05:00

Better late than never. ;-)

I missed the Sunday Funday post last week and I really don't want to make that a habit so this week I'm posting two versions of this week's game ... 3D Tic-Tac-Toe. Here are the instructions for the game pictured on the left (may take a little while to load) ...

The object of 3D Tic Tac Toe is to get four in a row horizontally, vertically or diagonally on one plane or across all four planes.

You are Red, the Computer is Blue. In the first game, you go first. In subsequent games, the loser goes first. If the game ends in a tie, whoever went first will do so again.

The moves of the game are notated on the right. Moves that threaten a win are noted with an *. The Computer plays a strong, but not unbeatable game. Good luck!!!

These are the instructions for the game on the right ...

This game is basically a 3D tic-tac-toe, except that the rules have been changed to keep the first player from winning all the time. Here you have to form two rows that meet at a right angle (in the shape of an "L").

Have Fun!!

Blogging on blogging

2006-05-28T17:05:00.000-05:00

I like this unit because it is more algebra than the previous unit that we have. Maybe, analyzing the word problem is giving me a hard time so far. However, more practice I think I can handle it on our test.

BLOG ON LOG

2006-05-28T14:40:00.000-05:00

The logarithms and exponents unit was a real challenge. Mainly because of the new concepts and new ways to look at fuctions in terms of exponents. It takes some time to get your head around it but eventually i get it. And I'm kinda confused in what ln is and wats the purpose of it.

Last Scribe of year

2006-05-27T21:33:00.000-05:00

Today, acually friday, we had a sub for both classes.

In the morning we worked in groups on a group assignment, about Logarithm word problems.
The question on the assignment was:

Sugar is put into a large quantitiy of water and the mixture is stirred. After 2 minutes 50% of the sugar had dissolved. How much longer will it take until 90% of the sugar has dissolved (Model the sugar dissolving as an exponential decay function.) ?

What we know :
Original = 100%
This is 100% because when we begin the experiment none of the sugar has dissolved, therefore there was still 100% in the mixture.
Final = 10%
This is 10% because we want to know how much time it will take 90% of sugar to become dissolved leaving 10% of the sugar left.
Life = 1/2
This is because we know that after every 2 minutes of the experiment 50% of the sugar will be dissolved which is also known as a half-life.

With that information we plug it into the equation:
0.1 = 1 ( 1/2) ^t/2

0.1 Represents the the final which is 10%
1 Represents the initial which is 100%
1/2 Represents the life which is 50%
2 Represents the amount of time it will take for the life to decay

0.1 = 1 ( 1/2) ^t/2
0.1 = (1/2) ^t/2
ln 0.1 = (t/2) (ln 1/2)
(ln 0.1) / (ln 1/2) = t/2
2 ((ln0.1) / (ln 1/2)) = t
6.6439 = t
It will take approximately 7 minutes for 90% of the sugar to become dissolved in water.

*NOTE: I am not sure if this is the right answer as we did not go over it in class, but it is what the majority of the class got so i'm assuimg it is.

After the group assignment we did some questions from out text book for the rest of the period.
Pg. 83 Questions 30 - 34

In the afternoon we just did questions in our text book for the period.
Pg. 119 Questions 1 - 33 Odd

That was all we did for friday's class. Last scribe of the year and I get a not so bad one =D. Anyway monday's scribe is Jefferson haha who better carry out our plan for next week =P.

B.O.B.

2006-05-26T18:50:00.000-05:00

LOGARITHMS . . . .umm I sort of get it. Long answer questions are somewhat complicated though because I don't know which equation to use and what to do. This unit was really short and so I found it hard to absorb all the log information. Hopefully we'll practice some long answer questions? Well there isn't much to say about this unit, hopefully the test won't be hard? haha I doubt it. Later math people.

=) =) =) =) =) =) =)

LAST SCRIBE OF THE CYCLE

2006-05-25T20:21:00.000-05:00

Today we had one class for today. We started off by writting down notes in our Math Dictionaries.

The Compound Interest Formula

A is the amount at theend of the interest period.

P (A.K.A. A_o) is the principle amount originally invested.

r is the percent rate of return. Written as a decimal.

n is the number of times the principle is compounded each year.

t is the time in years.

Interest Compounded Countinuously

A is the amount at the end of the investment period.

P is the principle amount originally invested.

r is the percent rate of return. Written as a decimal.

t is the time in years.

That was all the notes we had written down. Next, he wrote down some examples then problems.

Examples

C₁₄ has a half life of 5700 years.

Some bones that have been found only have 10% of C₁₄ that they should have. How old are the bones?

A colony of wasps has 800 members. Every 8 days the colony triples in size. How many wasps will there be in the colony in 3 weeks?

There will be 14306.

Solve

Radon has a half life of 3.8 days. How much of a 5g sample will remain after 4.2 days?

The population of a colony of pre-cal 40s students grew from 3x10⁵ to 4x10⁵ during the period between noon and 2pm.At what time will the population be at 6x10⁵?

Well that's all that we did for class today. Now the next scribe will be . . . . . . . ABRIEL muahahahhahahahahah ;D

DON'T FORGET TO DO YOUR HOMEWORK. . PAGE 82 #24-29

thats all folks

blogg on the logs

2006-05-25T01:14:00.000-05:00

this topic seemed pretty easy, the only thing i need to learn and pac down currently is a number to the power of a log, that stuff kind of doesnt "click" in my brain. other than that, i think i should do pretty good on the test, im aiming for success!

May 24th Scribe By Marcquin

2006-05-24T23:06:00.000-05:00

Well lets see here... Todays first class in morning we began with some questions.

Write as an expontial
a) log₂64 = 6
64 = 2⁶

b) log₇ 343 = 3
343 = 7³

Write as a logarithm
a) 3⁸ = 6561
log3⁶⁵⁶¹ = 8

b) 25^3/2 = 125
log₂₅125 = 3/2

Evaluate:
a) log₂4^3/2
log2(2²)3/2
log₂2³

b) 5log₅¹⁰ = 10

Solve:

log_t4 = 2
2² = t²
2 = t

Expand

log_{2 (m³ n⁵)^{1/2 = 3/2 log₂^{m+ 5/2} log₂ⁿ
bring down the 1/2, therefore

1/2 log₂ (m³ n⁵)

Remember "power of a power"

log₂ (m^3/2 n^5/2)
multiply the exponents
log₂m^3/2 + log₂n^5/2

Given: log₅ = 0.70 log₂ = 0.30
without using a calculator find log 0.08

Start of by saying 0.08 = 8/100
If you look at the number it can be simplified to 2/25

You must use a power to get a power which will get you 2/ 5²

Answer: log 0.08 = log (2/5²)

0.08 = log 2/5²
= log 2 - 2 log⁵
= 0.30 - 2 (0.70)
= 0.30 - 1.40
= - 1.10

Solve: log_ax + log_a(x-2) = log_a³

log_ax(x-2) = log_a³
x²-2x = 3
x²-2x-3 = 0
(x-3) (x+1) = 0
x=3 Accept
x=-1 Reject

We also had a talk about the population question in class. I am sorry guys because I lost the notes for it.

Journal Notes

EXPONENTS AND LOGARITHMS

Exponential Function: Any function where the variable is an exponent written in the form}}_{^{f(x) = abTHE ROLE OF PARAMETER A
a determines the y-intercept of the function a.k.a. the initial value (i.e. x=0) of the function.
a<0>the graph is reflected in the y-axis

THE ROLE OF PARAMETER B
b is the base of the exponential function
b is also known as the multiplication factor.
b>1 the function is increasing. This is known as "exponential growth".
0the function is decreasing. This is known as "exponential decay".
LOGARITHMS
Definition: 1) A logarithm is a function that turns a power into an exponent.
2) The logarithmic function, log_ba = c, is the inverse of the exponential function b^c = a where b does not equal zero and a>0.
^{^{log_b a = c}}
^{^{b is the base}}
^{^{a is the argument}}
^{^{c is the logarithm}}
^{^{which means}}
^{^{b^c = a}}
^{^{b is the base}}
^{^{a is the power}}
^{^{c is the exponent}}
^{^Examples:}log₃⁸¹ = 4 means 3⁴ = 81
^{^{log₂³² = 5 means 2⁵ = 32}}
^{^{log₄ (1/16) = -2 means 4^-2 = 1/16}}
^{^{log₅^1/125 = -3 means 5^-3 = 1/125}}
^{^{log₁₀₀ = 2 means 10² = 100}}}}
^{^GRAPHICALLY}
^{Since the logarithm function is the inverse of the exponential function...}

Remember: A logarithm is an exponent!

LAWS OF LOGARITHMS
Product law: log_amn = log_am + log_an
Example: log₂(4 * 32) = log₂4 + log₂32

Quotient law: log_a(m/n) = log_am - log_an
Example: log₃(81/27) = log₃81 - log₃27

Power law: log_am^b = b log_am
Example: log₅25³ = 3 log₅²⁵

Change of base law: log_am = log_b^m all over log _b^a
Example: log₈⁴ = log₂⁴ all over log₂⁸

SPECIAL CASES:

1) log_aa = 1
2) log_a1 = 0
3) log_aa^x = x
4) _alog a ^x = x

The Common Logarithm
A logarithm function to the base 10 ( log₁₀x) is called a "common logarithm". This base is so frequently used that it is simply written as log x, without the base indicated.
When no base is indicated, base 10 is assumed.

THE NATURAL LOGARITHM

e = 2.718281828459...

e is a number that arises naturally in the study of exponential functions, particularly in the case of continuous exponential growth. The natural logarithm function is written as:

lnx i.e. log_ex = lnx

read as "el-en of x"

And thats the whole bottle of wax for today!

So I will end off with a joke, haha just joking guys. So the next scribe has to be jan. It was inevitable!

blogging logs..

2006-05-24T15:55:00.000-05:00

well this unit went by really fast.. and just like always theres always a test after. this unit is kind of complicating for me. i get mixed up with whats the power, base, exponent and so on. I'm not having to much trouble when it comes to the product rule, quotient rule and power law, but i do get confussed when there are alot of logs involved in the problem. when it comes to everything else i guess i'm okay with it.

MR. BOB for LOGS AND EXPONENTS

2006-05-23T22:09:00.000-05:00

this unit started goood, the problems we did in the first few days were simple to understand. I didn't find the problems difficult. Today's topic or section we looked over was a bit confusing. Maybe it was "that i forgot what a log is", lol but i do not think it was that. It was just the way that the log was used and exponent, until i saw the pattern. This unit was fun, because it gives me a different way to think and there are many ways to solve an equation. I find that is pretty similar to trig identities, the way that problems should be solved. A little more practise on the problems we did today and maybe using them in word problems or something and im good to go.

Logarithms; What Fun

2006-05-23T19:20:00.000-05:00

Can you sense my sarcasm? Oh, logarithms aren't so bad, really. I mean no sarcasm. We just need practice right? Well, what's a better way to start the week, than with a one period math class? We started class with eight questions given to us on the board; all to solve for x.

a) 2^x = 3
b) 4^3x = 2
c) u^x = v
d) a = bc^x
e) log₅x = -2
f) log₃x = π
g) y = log₄3x
h) v = log_bcx

ANSWERS:

a) x = log₂3 OR log3/log2

How we get log3/log2 :

2^x = 3
log2^x = log3
xlog2 = log3
x = log3/log2 [divide each side by log2]

b)The way I solved this problem was to change 4 as a power of 2.

2^2(3x) = 2
2(3x) = 1
6x = 1
x = 1/6

Mr. K, then showed us other ways to solve b.

log₄2 = 3x
(1/3)log₄2 = 3x(1/3)
(1/3)log₄2 = x
log₄2^(1/3) = x
log₄³√2 = x

OR

4^3x = 2
log4^3x = log2
3xlog4 = log2
(1/3)3x = log2/log4(1/3)
x = log2/3log4
x = log2/log4³

c) x = log_uv OR logv/logu [it's exactly like question a), but with letters instead of numbers]

d) a/b = c^x
log_c(a/b) = x
log_ca - log_cb = x

e) log5x = -2
x = 5^-²
x = 1/25

f) x = 3^π

g) y = log₄3x
log₄3x = y [I just switch sides because it seems less confusing for me with the x on the left side]
3x = 4^y
x = 4^y/3

h) logb^cx = v
cx = b^v
x = b^v/c

So then we were given 3 questions to simplify:

a) 7^log₇4
b) 2^log₂5
c) 4^{log₂^[2(log₂5)]}

So first, we always have to remember a logarithm is an exponent. Let's use a different example. If given log₂32, we'd know that the exponent is 5. 32 would be the result of 2(our base number) to the power of an exponent, and in this case it would be 5.

2⁵ = 32

Then we can say that log₂32 is just another way to say 5. So if we were given 2^log₂32 , it would be like saying 2⁵, which is 32.

2^log₂32 = 2⁵

= 32

So if given different numbers, like 3^log₃5, it would be stating that 3 to the power of a certain number, will result in 5.

3^x = 5

So, 3^x is a certain number itself, which is equivalent to log₃5 OR x, and will result in getting 5 in this given question.

If 3^x = 5 then log₃5 = x THEREFORE 3^log₃5 = 5 is the same as 3^x = 5

I think that's how it goes. I hope I explained that right, but i think I may have just confused you guys even more. If that's still a bit vague, you can always just distiguish the pattern. If the base number is equilvalent to the base number attached to log, then the bigger number [sorry, I don't know the proper terminology], will be your answer.

a^log_ab = b

EX. 6^log₆8 = 8

So, the answers would be:

a) 7^log₇4 = 4
b) 2^log₂5= 5

c) 4^{log₂^{[2^(log₂5)]}}= 4^log₂5 [2^log₂5 is just 5, using the pattern we

identified]

= (2²)^(log₂5) [we change 4 as a power of 2]

=(2^{2^log₂5})²

=2^log₂25

=25

After that, we were given:

CHANGE OF BASE LAW

log_am = log_cm/log_ca

EX. log₃7 = log7/log3

=ln7/ln3

Mr. K then gave us two questions to work on, but we only had time to answer one.

1. log (x+6) + log(x-6) = 2

log (x+6)(x-6) = 2

log(x²-36) = 2

10² = x²-36 [I'm not 100% sure if that's how they relate or not]

100 = x²-36

0 = x²-136

0 = (x-√136)(x+√136)

x = √136 [accept] AND x = -√136 [reject]

2. [We didn't get to finish this question]

1/2log_a(x+2) + 1/2log_a(x-2) = 2/3log_a27

Our homework is seven questions that's posted on the blog. Yay, I'm done my scribe! Marc, you're the next scribe. Have fun with covering two classes. =)

blogging logarhythms

2006-05-22T22:22:00.000-05:00

Ok sorry for the late blog. First things first using logs remember these

Use this with the quotient questions
Log_a(m/n)=log_am - log_an

Use this with the multiplication questions
Log_a(mn)= log_am+log_an

This one I'm not sure when to use it but it's tricky
cLog_am=log_am^a

express as a logarhythm of a single number or expression
a)log_a2+ log_a3
Log_a(mn)
=log_a(2*3)

b)log_a5-log_a4
Log_a(m/n)
=log_a(5/4)

c)3logLog_a2
log_am^a
log_a3²

d)-Log_a(1/6)
log_am^a
log_a(1/6)^-1

e)1/2log_bm + 1/2 log_bn
Log_a(mn)
There are two ways to do this you can either put it as log_bÖ(mn) or
Log_bm^1/2n^1/2
=log_bmn^1/2

f)1/2(log_bx-log_b4)
Log_a(m/n)
=log_bÖ(x/4)

for these ones let C=Log₃(10) and let d=Log₃(5) and express in terms of C and D which means that when you simplify it it and it looks like either C or D then input it in for example Log₃(10)+Log₃(5) then replace it and you get C+D if you don't get it I'm sorry I don't really explain well but here are the questions you'll get and answers you'll see what I mean...

a)Log₃250
=log₃^(5²*10)
=log₃5²+log₃10
=2log₃5+log₃10
+2D+C

b)log₃log²
=log₃10/5
=log₃10-log₃5
=C-D

c)log₃Ö(5)
=log₃(5)^1/2
=1/2log₃5

d)log₃1/100
=-2log₃10
=log₃(1/100²)
=log₃1-log₃10²

it just says to expand so these questions
a)log₂m⁵n⁴
=5log₂m+4log₂n

b)log₂(mn)³
=3log₂m+3log₂n

c)log₂ ³Ö(m²n)
2/2log₂m+1/3log₂n

d)log₂ (Ö(m/n³))
=1/2log₂m-3/2log₂n

solve these questions plus mr.K used something called the anti-log as shown on question A it gets rid of the log
a)log_ax=log_aa+log_a5
=anti-log_a(log_ax)=anti-log_a(log_a45)
x=45

b)log_ax=log_a ³+log_a⁵
=log_ax=log_a ³Ö(9)²+log_a²
=anti-log_ax=anti-log_a ³Ö(9²)+ log_a²
=log₅₄
x=54

c)log_b(x-+3)=log_b8-log_b2
=anti-log_b(x-+3)=anti-log_b8-log_b2
x+3=4
x=1

This is where my notes get messy and I'm really unsure which is question and which is answer so I'll put it down as I see it.

d)log_b(x²+7)=2/3log_b64
log_b(x²+7)=2/3log_b64^(2/3)
anti-log_b(x²+7)=anti-log_b16
x²+7=16
x²=9
x=-3, 3

Also another note you can check your answer by filling in for x

e)log_bx-log_b(x-5)=log_a
log_b(x/(x-5))=log_a
anti-log_b(x/(x-5))=anti-log_a
x=6(x-5)
x=6x-30
x=5

f)log_a(3x+5)-log_a(x+5)=log_a8
anti-log_a(3x+5)-anti-log_a(x+5)=anti-log_a8
2x+5=8x+40
-35=5x
-7=x

I'm not sure if these are from the questions he gave us I just can't remember, but from what I see I think you would have to change the whole number into a logarthym and then you use the anti-log to find x

a)log₂(x²)=4
log₂(x²)=log₂16
anti-log₂(x²)=anti-log₂16
x=-5, 5

b)log₃6(x+2)+log₃6=3
log₃(6x+12)=3
log₃(6x+12)=log₃27
anti-log₃(6x+12)=anti-log₃27
x=5/2

c)log₆(x+1)+log₆=1
log₆(x+1)+log₆x=log₀1
anti-log₆(x+1)+anti-log₆x=anti-log₀1
\ no solution

Sorry for the long wait I just felt like doing the blog on a monday well it's a tuesday now but I started on monday If you see any mistakes feel free to tell me cuz I did it pretty late there's bound to be lot's of mistakes though, especially the last question. For the scribe umm... does anyone remember beat it by micheal jackson remember near the end where everyone crowded these two thugs and those two thugs had a weapon in one hand while the free hand was holding on to the other opponets free hand and the game was to hurt the other guy... well since there is marc and jessica left I believe they should do that to see who would be the scribe...winner doesn't have to do be scribe

Lemon square!!! Scribe post!!!

2006-05-17T15:56:00.001-05:00

Hi! guyzzz... We started the day with our terrifying test(just kidding). And finally Mr.K brought the lemon square, by the way its good Mr.K and can I have the recipe for that one.

In the afternoon, we started an example;

Evaluate the expression:
1.)

2.

3.

Then,Mark asked what if the example were given like this:

a.) 3² + 3⁵

b.) 2⁷ - 2³

Mr.K , explained that there are no such rules that can fit in these such question because the first 2 example are implied within the property of multiplication and division.

Remember this following rules:

There is also an example where the 2 rules are join together like:

Well, I guessed that's it for me, by the way our homework is on page 106, numbers 1-42 only the even numbers. The next scribe will be Lerwyne.

scribe post (post # 200)

2006-05-16T20:30:00.000-05:00

although i didn't exactly finish this post as the 200th post, based on the post arrangement i'm still the 200th blogger here! well, i'm aldridge and i'm the scribe for today. i'm not looking forward joining the scribe's hall of fame for this scribe post cause this topic is really confusing to me and i don't get some of it. well today is a 1 period class. everybody seems to be earlier than the bell today and most of the class is passing around 3 bubble teas to be tasted. the bell rang twice and after a few moment Abdi enters the door. seems like no one notice him cause everyone is busy doing something except for Van who immediately put Abdi in the board after noticing him. Hopefully, Mr. K will bring the lemon squares tomorrow.
We did some review about probability today. the results form our quizstar quiz are already in. Mr. K discuss to us all of the question from that quiz. here they are:

these are the top 3 questions where most of us who take the quiz got it wrong. just click on the image to make it bigger.
here are the other 7 questions with their answers:

out of 16 students who do the quiz, the class got a total average mark of 72%, not bad for this unit.

sorry if this post took me long to post up. some of the statements I made here are already past tense but since the date posted is still may 16, its better to leave those statements in present and future tense. the next scribe will be emile.
I hope this is my last scribe post. haha... jk..

Homework Assignment for May 16

2006-05-16T19:05:00.000-05:00

Here is your homework assignment for tonight (20 Questions):

Questions Part 1 of 2
Questions Part 2 of 2

Here are the answers to the assignment:

Answers Part 1 of 3
Answers Part 2 of 3
Answers Part 3 of 3

This is post number 199 on our blog. Whoever posts next will write post number 200. I wonder what it will be? ;-)

Scribe#55 : Review and Brief Introduction To Logirhythm

2006-05-15T18:23:00.000-05:00

Okay, just because I wanted to look good, I spend 5 minutes counting what number scribe we're on. Apparently its 55. That's pretty cool. Anyways, I'm Van, and I'm gonna do one kick @55 scribe. (Sorry if that's offensive Mr. K).

Today was a 2 period day. Mr. K was lonely at the front. People slowly etched away, especially Sissyphus (That has to be a typo...). No one felt like moving, so Mr. K went on with the class. ALSO, Mr. K forgot about our lemon squares! The 80% probability failed. We gotta get em tomorrow!, hopefully, if Mr.K sees this scribe in time, he'll remember to make them right away.

So Mr. K posted the online quiz a LITTLE late, so it should end tonight at 11:59:59. So, get those done people! Mr. K mentioned about, all the BoB's(hehe I invented that) state that, most people have trouble with the poker probability. So we started doing 3 practice problems.

1. Find the probability of getting... *BOOM* *Mr.K turns around and looks at Calvin*

Calvin: "SO CLOSE!, I swear, that hole puncher was there as a trap."

Calvin failed to sneak into the room late while Mr.K wasn't looking. Now he owes him 3 bubble teas. Just thought I'd mention that.

1. Find the probability of getting 2 pair in a 5 card poker hand

A two pair is two cards of one rank and another two cards of another rank.

Ex.

2. Find the probability of getting exactly 5 spades and 4 hearts in a 13 card bridge hand.

Okay, so what in the heck is bridge? I have no clue. I've never seen anyone play it, so let's post the rules for the heck of it, in case anyone wants to try.

http://www.pagat.com/boston/bridge.html

Find the probability of getting a "Full house" in a 5 card poker hand.

A full house, or full boat, contains a set (three) cards of one rank and a pair of another rank.

So those were the practice questions, and the test will be on Wednesday MORNING. Get ready guys.

Sketch Each Pair Of Graphs On The Same Cartesian Plane

Mr. K: Start by gratch...

Mr. K said gratch. Definition: Mr. K's mistake of a new word. Appears to be a combination of "graph" and "sketch"

So, I probably should've wrote what he was trying to say about exponents and powers. I'll update this again and ask him about the language. For now, here's the questions and answers

All I know is that, whatever the log base(under the log), the exponent(the # after =), is what I'm looking for, to make the power(beside the log)

blogging

2006-05-15T00:42:00.000-05:00

so yea I think I got it, I said I think I get it I'm not exactly sure if I trully got it, but still I understand it alittle bit but if it came down to the test I'll probably forget it hahaha. The problems I don't understand is the word problems involving the tree, the words like trick you sometimes just when I get it, it turns this corner and cofuses me hahaha. ohh and sorry bout the diffirent coulors I jsut felt bored

blog post

2006-05-15T00:17:00.000-05:00

well, our unit, "probability" is really hard for me! this is the hardest unit for me now! probably because i already forgot about permutations, combinations and binomial theorem! i think i need to review again those parts in our past unit. this probability thing should be easy for me cause i already took this at my statistics class and algebra(just for fun exercises in our class). i'm still trying to remember the easier way for doing word problems in this unit that i learned about two years ago. i hope i can remember it before our test cause i guess the probability that i'll pass this test is low. :-(

blogging for the test!

2006-05-14T22:20:00.000-05:00

This unit was pretty short and way easier than the last unit. The hardest thing I would say about this unit is the questions dealing with a standard deck of cards. It just so hard to get my head to understand all those numbers in the deck. The easiest thing I found in this unit is using the tree diagrams to help solve the problem, it really helped. Well that's about it, oh and good luck everyone for the test!

blogging on blog

2006-05-14T21:02:00.000-05:00

I actually don't know what the test is about but, from what we do in class, it's seems pretty challenging. it's hard to grasp. When I first read the problem, it's just words and takes a while to actually get what the question is asking. Like what to use, the Choose thing or something else, what do first, what to do last. it's confusing. =S

blogging on blogging

2006-05-14T20:06:00.000-05:00

This unit is sort of easy. There is just some trouble when it comes to organizing the problem. For example when you have to find the sample space. I just did the online quiz and I started to understand how to do the questions. Another place that i have trouble on is when the notation GIVEN THAT is involved. I guess i just have to be more open minded.

Blogging for probabilty on the blog

2006-05-14T17:34:00.000-05:00

My blog is kinda late, but here it is. This unit realy does seem similar to the previos unit, counting. I would also say that this unit a little more easier compared to counting, because we do not have to deal with circles. This unit also helped me to better understand the use of the compliment. Some questions are are really tough to do, but it's fun since they are challenging. I have to be careful for the wording of some questions, because we have learned that "and" and "or" mean different things for probabilty questions. Other than that i don't think i had much problem with this unit. Again, practise is the key, i think to understand this unit and really diggin deep into what a problem asks for.

Probability Online Review Quiz

2006-05-13T11:08:00.000-05:00

I had some trouble with Quizstar yesterday. The quiz went online this morning. I've extended the deadline to midnight Monday night. The unit test will be on Tuesday or Wednesday. We'll talk about it in class on Monday.

You can write the quiz here. You have 30 minutes to complete it. If you haven't yet registered at QuizStar then when you get to the site follow these instructions:

Click on the big yellow "Sign Up" arrow.

Use only your first name and last initial as indicated.

Pick a username that will allow me to easily identify you, i.e. first name and last initial.

Make up any password you like.

Click on [Register] then [Search] by teacher's name (kuropatwa) and you'll find me.

Click on the box next to Pre-Cal 40S and then [Register].

Follow the instructions on the screen.

Actually, if you read each page carefully, you'll see that the sign up process is very straight forward and self explanatory. If you hit any snags email me and we'll sort it out together.

Remember, the quiz is timed. You'll only have 30 minutes to complete it once you've begun. It consists of 10 multiple choice questions (1 mark each). DON'T PANIC. Take your time. I know you'll all do well. ;-)

Do Your Best!

OMG 10 AMAZING SCRIBES IN A ROWW

2006-05-12T23:49:00.000-05:00

Oh my god Jacky that was an AWSOME scribe and thanx for choosing me the lucky scribe for today:) I would also like to say to the last 9 scribes in a row they did an amazing and an out standing job on there scribes.
Hay class mates i hope that you are starting to study for this units TEST. This unit was short unlike the other ones. Okay today was a one period class and as always it starts at 1:20 pm and if you were not in Mr.K class by that time you must bey him Ice Tea. We got alot of flavours covered alrady but we want to cover them ALL so keep on working on that class mates :) And today we just did a pre-test.

The pre-test is as fallows
Calculators are allowed.

Question:
1) Twelve people, including you, are members of a choir. The choir director is going to choose three members to attend a workshop. The probability you and two other members will be chosen is:
a) 1/4 b)3/10 c)1/12 d)1/10
Solution: Ther are 3 to be chosen out of 12 people 3/12 reduce it = 1/4 Therefore, the answere you must choose is (a).

Question:
2) Rex is playing a guessing game. The probability he will guss each question correctis 0.3. What is the probability he will guess exactly 5 out of 10 questions correct?
a)0.15 b)0.10 c)0.29 d)0.80

Solution: (C)= Correct answers (W)= Wrong answers p(c)=0.3 CCCCC WWWWW (10C5)(0.3)^8(0.7)^5 =0.10 Therfore, the one you choose should of been (b)
Question:
3) The serial number of a $10 bill contains 8 digits. If your $10 bill contains the digit 7 at least once, you win a prize. What is the probability that you'r $10 bill will win?

Long Answer Questions
4) Jhon takes the bus to school on two days, and on the other three days he walks. If he takes the bus he is late 10% of the time, but if he walks he is late 30% of the time.
a) What is the probability that John arrives on time?

Solution: On time:p(Bo)+p(Wo) =0.36+0.42 =0.78 (100) =78%

b) If jhon was late, what is the probability that he took the bus? solution:
P(Bl/l)= p(Bl)/p(Bl)+P(Wl) =0.04/(0.04)+(0.18) =(0.4)/(0.22) =0.1818 (100) =18%

And the Next Lucky scribe is .......i need a drum roll :) I can't dacide so i put the names in a hat and i picked one out and the next scribe is Van. It was the only way i could be fair and after all this is the end of the probability unit. :P :P

Blogging on blog

2006-05-12T10:38:00.000-05:00

Hay it's time for (BOB)Blogging on blog. First thing I want to congratulate the 10 AWSOME scribes for there hard work and accomplishment. There post's are really good for studying notes and i would Definitely recommend to my class mates to read the posts while studying for the test. I am having problems with the card problems they confuse me and I don’t like them because I don’t play cards (i am serious). That's the most think I am having problems with and i need more practice on. well i wish every body good luck on the test and there is just 2 more units for the course to end (time really did go by so fast)

Scribe: Introduction to Exponents and Logarithms

2006-05-12T02:10:00.000-05:00

Well today was a two day, and lucky me, I got chosen for it fair and square.

Let's see, well the morning started off with Mr. K telling us things to know and expect, when doing something for the first time. Like riding a bike, or in this case dealing with math. Since you're doing it for the first time, you can't expect to learn it right a way, you're going to have to work at it. Education is about pushing yourself, to that level where you're not just satisfied, but to that level of excellence.

Then Mr. K put us into groups, where we had to solve probability questions.

1. Two soccer teams, A and B, have a kickoff to see which team wins the game. The teams take turns to attempt to score a goal. The first team to score wins. The probability of team A winning with one kick is 0.20. The probability of team B winning with one kick is 0.25.

a) Suppose team A kicks first. What is the probability that team B wins on its first kick?
First Kick
AN*BG
(0.8)(0.25)= 0.2
=20%

b) Suppose team A kicks first. What is the probability that Team A wins on it's third kick?
1^st, 2^nd, 3^rd kick
(AN*BN)(AN*BN)*(AG)
(0.8*0.75)(0.8*0.75)(0.2)=0.072
=7.2 %

2. Five dimes and five nickels are arranged in 3 boxesas follows: Box 1 has 2 nickles and a dime, Box 2 has 2 nickles and a dime, Box 3 has 3 dimes and a nickel. A box is randomly selected, then a coin is randomly selected from that box.

What is the probability that the coin will be a nickel?
P(1N)= 1/3 * 2/3
= 2/9

P(2N)= 1/3* 2/3
= 2/9

P(3N)=1/3* 1/4
=1/12

P(N) = P(1N) + P(2N) + P(3N)
= 2/9 + 2/9 + 1/12
=8/36 + 8/36 + 3/36
=19/36

3. A hand of 13 cards is dealt from a shuffled deck of 52 cards. What is the probability that there are:

a) exactly 5 spades?

b) at Most 5 spades?

Then comes our introduction to the new unit in the afternoon; Exponents and Logarithms. These are some basic stuff we should already know about exponents:

Then Mr. K gave us four numbers on the board. We had to write each of them 4 different ways using only exponents.

***A number can be written in many different ways.

These were the numbers

4
(1/2)^-2
(16)^1/2
(1/4)^-1
(1/8)^-2/3

1/2
(2)^-1
(4)^-1/2
(1/4)^1/2
(8)^-1/3

3
(1/3)^-1
(9)^1/2
(1/9)^-1/2
(27)^1/3

3/4
(4/3)^-1
(9/16)^1/2
(16/9)^-1/2(27/64)^1/3

***Any base must be in reduced form!!

Of course there is a pattern into figuring the different ways of writing a number. Using the basic stuff to find the first couple of ways, then you can use that to help you find the pattern.

Then Mr. K gave us another type of question on the board that had to deal with a variable in the exponent.

Here's another question Mr. K gave us, that I should warn you about.
It is common that people will do this in the next question:

3(5^x+1)= 15
15^x+1= 15

DO NOT DO THAT!!!

Because of BEDMAS, exponents should go first before multiplying,
YOU CAN'T RAISE 5 TO THE EXPONENT X +1

So it is solved like this,
3(5^x+1)= 15
5^x+1= 5
x + 1 = 1
x= 0

Double check:
3(5^x+1)= 15
3(5⁽⁰⁾⁺¹)= 15
3(5¹)= 15
3(5)= 15
15=15

Mr. K gave us many practice questions, so here they are along with the answers:

Q.27^x= 9^2x-1
^{^{A. x=2}}

Q.4^2x-1= 64
A. x=2

Q. 32^3x-2= 64
A. x= 16/15

Q. 4^8x= 1/16
A. x= -1/4

Q. 6^3x-6=1
A. x=2

Q. 5^4-x= 1/5
A. x= 5

Q. 3^2x-1 + 1 = 2
A. 1/2

I just want to say, very sorry guys for the late scribe, but i FINALLY FINISHED.

Okay well thats about it for the two classes today, for the people who weren't in class there wasn't any dictionary notes for today. Oh and there's a pretest tomorrow. Also Homework was page 89 1-16 and 24.

The lucky scribe for tomorrow is Zaenab.=)

hi bob!

2006-05-12T00:32:00.000-05:00

im getting probability, a few more card questions and it should beable to be packed down. im feeling okay about the topic, i need to feel the urge to " own everyone " haha ill get there soon before the test i hope =)

BoB was here ?

2006-05-11T23:53:00.000-05:00

well this topic here, probability, i acually feel confident about it. Scared at the same time, why you ask because when i'm confident i get bad results so hopefully this one is different becuase i'm going with full confidence on this one. like what mr. k said let's go for the 'i'm on top of the world' attitude. all this confidence has gotten me wired and excited for what else is coming so i'm hoping for better results in the future.

blogging on blogging

2006-05-11T22:48:00.000-05:00

This unit went by so fast, I think I understand it all pretty well. The tree diagrams and the probability questions are alright. We've worked with the questions repeatedly, so I think I'll do okay on the test. The questions aren't very hard as Mr. K had mentioned, they're just very long, which I don't like. I think that's all I have to say for this unit. Bye math people.

Blogging on blogging

2006-05-11T21:56:00.000-05:00

Well for me, probability is a lot easier than the counting.The only thing that I have a problem with is when dealing with the combinatorics permutation with probability. The rest I think is I can handle, as far as I know.Just few more example and practices I think I get it.

bloggin

2006-05-11T21:36:00.000-05:00

wow this unit ended very quickly, and that means another test !! hmm... what am i having trouble with? well this unit was alright for me. The three diagrams really do help alot. I guess the one problem that i do have probably has to deal with finding the sample space. When it comes to everything else I'm doing okay.

YO YO YO BoB is here

2006-05-11T21:34:00.000-05:00

Alright guys, I think this unit had its ups and downs. I liked parts of it and found the rest of the material difficult to grasp. If you give me a question involving a deck of cards you might as well give me that zero out of 4 because I believe that this is the most hardest thing to grasp. So.... I will need help on everything in this unit!

Everyone must study in order to pass. You must aim "to kick butt" and not just to aim "to get by".

BoB

2006-05-11T21:19:00.000-05:00

I think this unit is a lil bit easier than the last one but its complicating especially how or what to start with in a probability question. Or the ending of a question will be harder. example if a question is got a, b, c, and d. a and d will be hard to get. But this needs lots of practice. newayz u guys study and good luck on monday's test.

Sisyphus' Bloggon the Blog 6

2006-05-11T20:47:00.000-05:00

Like I said in my previous 'BoB' post, counting and probability is my weakest unit in math. However, now I do think I am getting stronger at it because of the repitition. The problem that lies within me, I'd probably say that it is my ability to read questions 'wrongly.' My 'spelling errors,' as Mr.K would put it. As well as the habit of second guessing myself. But I do understand the concepts, and the group work today helped proved that to myself (e.g. adding 2/9 + 2/9 + 1/12 to equal 11/36... sorry to you guys in my group, really). Mistakes are made to be learned from, it just takes me about twice to get it. I'm looking foward to the pre-test tomorrow, I really am. Good luck to you guys, and remember to Sisyphus.

The Probability BoB

2006-05-11T18:40:00.000-05:00

Wow this unit passed by so fast. Overall this unit was alright. Like I know what to do every type of question but it's not as clear to me. Like when to make a tree diagram. But overall this unit wasn't all that difficult because it's very similiar to the counting unit.

BoB

2006-05-11T17:46:00.000-05:00

Ohhh yeah, I've been recognized to have invented the BoB. Anyways, I'm not having any real troubles with the probability. Just taking a while to get used to. Maybe a little bit more practice and I'll be ready for the test.

probability blogging on blogging

2006-05-11T16:43:00.000-05:00

YES!!! It's the end of the unit! I didn't really enjoy probability. I found it hard through out the whole unit. Everyone has their weakness, so probability or any other problem solving questions I'd have to say are my weaknessess.

The only probability questions I liked to solve were the conditional probabilities. They were easy for me to understand and I know that using tree diagrams really help for those types of questions (and I really like using tree diagrams). But as far as the other probability questions I still having a hard time solving them. I think I can solve the questions but for me to solve them I would need more time than what is required. The questions I really don't like are the card problems because those questions always have at least one restriction.

Well that's all... good luck everyone in the test! and STUDY!

Bogging on blogging

2006-05-10T20:36:00.000-05:00

I am really have a hard time mastering probability. The questions I read seems like a whole new language to me. I don't know what else to say. I need work understanding the problem. I think I understand how to calculate the probability using "and and "or" but when I try to solve a question that has those, I go blank. I understand how to use the tree diagram. But sometimes I have a hard time building the tree from questions that I find difficult. Every word problem I try to solve, it always gives a hard time understanding it. I just need a lot of practice.

2006-05-10T17:56:00.000-05:00

Eight Hall of Famers in a row...Pressures on =S. Sorry if I kill the winning streak.

Mr. Kuropatwa was late today..That means l e m o n s q u a r e s ..yayy =) Can't wait. There's a new empress in town, yes Charlene did an awesome job. Anyways there will be an upcoming online quiz, which consists of 10 questions, 30 minutes to write it. It'll probably activated over the weekend. The Probability TEST will be on MONDAY. You know what that means, blogging on blogging posts.

We started off class with some board questions:

A 25 member high school committee consists of 7 female teachers, 8 male teachers, 6 female students and 4 male students. Find the probability that a 5 member sub-committee:

A) consists of only males
B) consists of only teachers
C) is made up entirely of male teachers
D) contains at least one female student

A new medical test for Glaucoma is 97% accurate. Suppose that 0.6% of the population have Glaucoma. Find the probability that:

A) A randomly selected person will test negative

B) A person who tests negative has Glaucoma

C) A person who tests positive does not have Glaucoma

Solutions:

P(N)= P(GN) + P(HN)

P(N)= 0.00018 + 0.96418

P(N)= 0.9644 or 99.44%

B) P(GN) = P(GN)/( P(GN)+P(HN) )

P(GN)= 0.00018/(0.00018+0.96418)

P(GN)= 0.0187%

C) P(HP)= P(HP)/( P(HP)+P(GP) )

P(HP)= 0.02982/(0.02982+0.00582)

P(HP)= 0.582%

After completing these board questions we received a worksheet and got into groups.

Here are some questions on the worksheet:

The teacher of a physical education class has two bins of balls. Bin A has 4 soccer balls and 6 volleyballs. Bin B has 2 soccer balls and 8 volleyballs. The chart below shows the probability of a student selecting either bin, and then a ball within that bin.

A) Determine the probability that a student will select a volleyball.

B) Given that the student selected a soccer ball, determine the probability it came from Bin B.

C) If the students are permitted to use bin A, what is the probability that a student will select exactly 2 soccer balls, and 1 volleyball if three balls are selected?

Solutions:

A) P(V) = P(AV) + P(BV) = 0.72 or 72%

B) Didn't quite get to finish because class had ended.

C) 10C1 + 10C2/10C3 = 0.4583

Well class ended before we got to finish the worksheet and check the answers. There was nothing for the dictionary today but don't forget to do blogging on blogging post. Homework, textbook page 396, questions 1-13.

next scribe: R a y m o n d (my little sister chose the scribe because I could not decide haha.)

Anyways this was a lot of fun but now I'm done so see you later.

=) =) =) =) =) =) =) =)

Scribe post : the end of probability

2006-05-09T16:48:00.000-05:00

well first all I would like to "thank" Abriel for picking me. wow my class is on a roll here..7 hall of famers in a row. CONGRATS! Now all the pressure is on me loL:( better make this good.

today was a two period day and was also the day we wrapped up the probability unit. In the first period class we went straight to our groups from yesterday and finished working on the problems.

#2) From a group of 8 men and 7 women, a committee of 6 is chosen. What is the probability that 3 men and 3 women will be chosen?

solution
₇C₃₈C₃ / ₁₅C₆

7C3 is the number of possible ways women can be chosen. The 7 represents the number of women and the 3 represents how many can be chosen.
8C3 is the number of possible ways men can be chosen. The number 8 represents the number of men and the 3 represents how many can be chosen.
15C6 is the number of possible ways with no restrictions. This is the sample space.

#3) Frankie take 1 language course, 1 math couse and 1 science course. The probability of Frankie passing a language course is 0.9, a math course is 0.7 and a science course is 0.6 frankies father picks one course at random and asks "Frankie did you pass?"

A) What is the probability that the correct answer is "Yes"?

solution

when answering this question you should start with making a tree diagram

M= math
F= French
S=science
P=pass
F=fail

In this question you are trying to figure out the probability of Frankie saying yes so you multiply the probabilities for the subjects and passing.Once you have those answers you add them together and you get you answer for the question.

P(MP) + P(FP) + P(SP)

= 7/30 + 4/30 + 6/30

= 11/15

B) If the correct answer is "Yes", what is the probability that the course was Math?

solution

P(MP) / P(MP) + P(FP) + P(SP)

= (7/30) / (22/30)

= 7/22

P(MP) + P(FP) + P(SP) is the sample space which we got from the previous question. Therefore this is the denominator. This is the sample space because this is all the possible ways Frankie can answer "yes" .
P(MP) is what you want to figure out so it is the numerator

#2) Eight students of different heights are seated randomly around a circular table. Find the probability that the two tallest students are sitting next to each other.

solution

step1) There are 8 students all together and they are going to sit in a circle. Because they are going to sit in a circle you have to set a reference point.

(8-1)! = 7!

step2) The question also states that the two tallest boy are going to sit together so you have to put them in a bag and count them as one. Because of this you are left with only 6 bags.

(7-1)! = 6! (you get the 7 from the previous step)

step 3) Now you have to figure out the number of ways the two people in the bag can sit.

step 4) now you figure out the sample space because this is a probability question. This is the number of all possible outcomes.

step 5) put it all together.

6!2! / 8!

= 1/28

After finishing the group work from yesterday we started on some other questions while still working in out groups.

Suppose a test for cancer is known to be 98% accurate (ie. The outcome of the test is correct 98% of the time). Suppose 0.5% of the population has cancer. What is the probability that a person who tests positive had cancer?

Suppose 1,000,000 randomly selected people are tested. 4 things can happen.

a person with cancer tests positive
a person with cancer test negative
a heathly person test positive
a healthy person test negative

#1a) How many of the tested people have cancer?

(number of people)(percent of people with cancer) = people with cancer

(1,000,000)(0.005) = 5000

b) How many do not have cancer?

(number of people) - (people with cancer) = people with no cancer

(1,000,000) - (5000) = 995,000

(percent)( number of people) = people with no cancer

(0.995)(1,000,000) = 995, 000

#2) the test is 98% accurate

a) how many people with cancer will test positive?

(people with cancer)(0.98) = accurate count

(5000)(0.98) = 4900

b) how many people with no cancer will test negative?

(people with cancer) - (people who will test positive) = people with no cancer who will test negative

(5000) - (4900) = 100

(percent that's not accurate)(people with cancer) = people with no cancer who will test negative

(0.02)(5000) = 100

#3a) How many healthy people will test positive?

(number of healthy people)(percent that's not accurate) = healthy people who will test positive

(995,000)(0.02) = 19,900

b) How many healthy people will test negative?

(number of healthy people)-(healthy people who will test positive) = healthy people will test negative

#4a) How many people tested positive for cancer?

#2a + #3a = people who tested positive for cancer

4900 + 19900 = 24800

b) How many of those people have cancer?

4900

c) What is the probability that a person who tests positive for cancer had cancer?

P(C(given that)P) = 4900 / ( 4900 + 19900 ) = 19.76

another solution

You can also make a tree diagram to solve this problem

To get these answers you just simply multiply along the branches

P(C(given that)P) = P(CP) / (P(CP) + P(HP)

=0.0049 / (0.0049+0.0199)

The afternoon started with some more probability questions...

A company makes computer chips, 70% of all chips come from factory A and 30% from factory B. In factory A 25% of the chips are defective. In factory B 10% are defective.

solution

The first thing that you should do is make a tree diagram

a) Suppose we do not know what factory a chip came from. What is the probability it is defective?

P(D) = P(AD) + P(BD)

=0.175 + 0.03

=0.205

=20.5%

In this question you are trying to look for the probability of the chips being defective. So in order to do this you add the defective probabilities for factory A and Factory B

B)You found a defective ship! What is the probability it came from factory A?

P(A(given that)D)= P(AD) / P(D)

=0.175 / 0.205

=85.37%

Since this is a probability question you need to have a sample space. In this situation we have already figured out the sample space in the previous question which is 20.5% or 0.205. We also need to have a number of favorable outcomes, In this case it is defective chips from factory A which is 0.175. To get the answer we divide the number of favourable outcomes by the sample space.

Dimes and Nickels are arranged in 3 boxes like this:

A box is randomly selected and a coin is chosen

The first thing that you should do is make a tree diagram.

a) The probability of a nickel.

P(N) = P(1N) + P(2N) + P(3N)

=1/9 + 1/9 + 1/4

=(4+4+9) / 36

= 17/32

In this question we just add up all the probabilities for getting nickels.

b) If a dime was chosen, what is the probability it came from box 3?

P(3(given that)D) = P(3D) / (P(1D)+P(2D)+P93D)

= (1/12) / (2/9 +2/9+ 1/12)

=3/19

Since this is a probability question, we need to have a sample space. In this case the sample space is the probability of all dimes in all three boxes. The number of favorable outcome is the probability of chosing a dime from box three. So to get the answer you just divide the number of favourable outcomes by the sample space.

Well I guess that about wraps up my scribe. And oh yeah I almost forgot the next scribe..

ANH

A Question For All The Scribes

2006-05-08T22:36:00.000-05:00

I found this comment on Manny's Scribe Post below. It's addressed to all the scribes in our class. That means all of you. Let's work together on this and teach others what you've learned. (I'm so proud of all of you!)

Congratulations to all the scribes - you've been doing some excellent work lately (and maybe helping me convince the math teachers at my school to try having scribe posts).

I have a question for all the scribes: I was wondering what software you are using to generate the math symbols, equations, etc. in your scribe posts? I can tell that you are creating it and saving it as an image, then uploading the image to Blogger, but not what software you are using. About the only thing that we have (that I know of) that could do something similar would be using the equation editor in PowerPoint, and then saving the slide (or just the equation) as an image, but that's pretty clunky. Do you have any software you can recommend (preferably free)?

Also, I notice you often use tables to put borders around items - are you generating the HTML by hand for that, or using something else?

Any ideas you are willing to share would be appreciated. Keep up the good work - you are setting a great example not only for other students around the world, but for their teachers.

This comment is from Karl Fisch. He's the Director of Technology at Arapahoe High School in Centennial, Colorado. Now you guys are really teaching the teachers! Cool.

Scribe ? Whos' Next ?

2006-05-08T19:39:00.000-05:00

So it's finally my turn to be scribe. Hmm.. this could be my last time being scribe before the exam too. The probability of that happening is very high as there seems to be a lot of people left for scribing so not only do I have to do a GREAT scribe and dethrown manny but also finish the year with a BANG! Anyway on to the scribe.

Today we only had a single class which began with the usual, Mr. K finishing his lunch while writing questions on the board.

1) A Pre-cal 40S studen takes a 20 question multiple choice test with four possible answers to each question. If she knows the answers to 12 of the problems, what is the probability of her scoring 90% on the test by guessing on the remaining questions ?

Well to start this question off we know that there are 8 questions remaining in the test, and in order for her to gain a 90% she MUST answer any of the next six questions right. The 'word' that we create with that information would be as follows : RRRRRRWW; were the R represents a right answer and the W representing a wrong answer. The catch towards this is assuming that she gets the next 6 questions right and the last 2 wrong, but having six right and two wrong can happen in many different ways, therefore we use the choice formula to figure out how many different ways it can happen :

₈C₆ = 8! / 6!2!

The eight is the number of questions remaining while the 6 is the number of questions she must answer right and the 2 is the number of questions she is allowed to be wrong.

Back to the word : RRRRRRWW

Since we also know the student has 1/4 chance of getting the question right and 3/4 getting a question wrong. It is 1/4 to get an answer right because there are four possible solutions to a question and only one is right our of three and visa versa for wrong answers.

From that and the word we know notice that her chances of getting 6 questions right is (1/4)⁶ and her chances of getting 2 questions wrong is (3/4)²

If we take the number of combinations she could get 6 right and 2 wrong (₈C₆) and the number of times she can get 6 right ((1/4)⁶) and the number of time she can get 2 wrong ((3/4)²), and multiply it all together we get the probablilty she will get 90% on her test :

₈C₆(1/4)⁶(3/4)²
= 28 (1/4096)(9/16)
= 0.38 %

2) A piggy bank contains 2 quarters, 3 dimes and 5 nickels. When you turn it over and shake it any coin can fall out. What is the probability of shaking out two coins that total more than 30 cents ?

Explainations are all on the image itself, if still confused feel free to leave a comment and either me or someone else will elaborate for you.

^{3) A family has 5 children. If the birth of a boy or girl is equally likely, what is the probability that the family consists of}
^{a) Exactly 2 girls}
^{b) at least 2 girls}

^{a) For this half of the question, is exactly like the first question I had explained. First we find the number of combinations for 2 girls, ₅C₂. The 5 from the number of children and the 2 from then number of girls you want. From there you then see what the ratio from girl to boy is which is (1/2)² to (1/2)³. The 1/2² is the probability of the number of girls and the 1/2³ is the probability of boys. Then we just multiply together like before :}

^{₅C₂(1/2)² (1/2)³}
^{= 10 (1/4) ( 1/8)}
^{= 10 ( 1/32)}
^{= 10 / 32}
^{= 31.25 %}

^{b) This question is easier done by finding the compliment. The compliment of having at least 2 girls is having 5 boys or 4 boys and 1 girl.}

^{5 Boys :}
^{P(oG) = 5! / 5! (1/2)⁵}
^{= 1 / 32}

^{P(1G) = 5!/4!(1/2)⁴(1/2)¹}
^{= 5 / 32}

^{With the answers from the compliemtn we then subtract one from the sum of the compliment.}

^{P(At least 2G) = 1 - (1/32 + 5/32)}
^{= 1- (6/32)}
^{= 32/32 - 6/32}
^{= 26/32}
^=81.25%

^{From there we got grouped together to work on a worksheet. Unfortunately we only had time to do the first question. I do not remember what the question was as Mark took it with him but I remember some of it.}

^{7 girls and 2 boys sitting at a round table. what is the probability of the 2 boys sitting together ?}

^7!2!/8!
^{= 2/8}
^{= 1/4}

^{The 7! is from the number of combinations you can arrange the girls, and the 2! is the number of combinations you can arrange the boys. The 8! is the total number of possibilities of both sexes. Notice how the total number of possibilities is 8! not 9!, this is because we are dealing with a circle here. One person has to be the reference point as they can sit anywhere therefore we subtract one as earlier discribed in an earlier scribe.}

FINALLY done, three hours of work whoa. VERY frustrated as blogger decided to delete my other finished scribe, and when I clicked on recover post it just recovered my draft *curse you. Very annoyed about that wow I sound mad and yeah I am but i'll let it steam out.

BTW Charlene, it's all yours tomorrow ;P

Kakuro Sunday

2006-05-07T11:11:00.000-05:00

'A Kakuro consists of a playing area of filled and empty cells similar to a crossword puzzle. Some black cells contain a diagonal slash from top left to bottom right with numbers in them, called “the clues”. A number in the top right corner relates to an “across” clue and one in the bottom left a “down” clue.

The object of a Kakuro is to insert digits from 1-9 into the white cells to total the clue associated with it. However no digit can be duplicated in an entry. For example the total 6 you could have 1 & 5, 2 & 4 but not 3 & 3. Sound simple? Be warned it gets hard and is as addictive as Sudoku.'

Click here for more Kakuros.

(Thanks again to Think Again!)

Scribe Post: Probability is 1 out of 2 times fun for me

2006-05-05T23:34:00.000-05:00

Hello everybody, my name is Manny and I’m your scribe for today courtesy of the newly famed Teddie. Today was pretty straight forward, it was a 2 period. The big counting test in the morning, followed by lots-a-notes in the afternoon. The test was pretty difficult, and I realized few questions I solved incorrectly as soon as I walked out of the door. It really is a bummer. I hoped you guys had better luck with it than I did. Anyway, I‘ll try to live up to you admired hall of famers. And here I go…!

In the afternoon class, we were given a huge amount of notes to absorb, with the explanation behind it of course. Probability is pretty tough for me for some reason. I’ll try to break it down, and explain it into more easier terms to make the notes more understanding by throwing in some words and sidenotes.

Let’s get started with some terminology of probability.

Certain Event: An event whose probability is 1.

Impossible Event: An event whose probability is 0.

Important: Probability is always a number between 0 and 1.

Complimentary Event: The compliment of E is E′, the set of outcomes where E does not occur (sidenote: the ′means prime)

If P(E) = a then P(E′) = 1 - a

Example:
The probability of drawing a spade from a standard deck of cards is

P(S)	= 13 / 52
	= 1 / 4

(sidenote: 13 = # face values in a standard card deck of 52)

The probability of not a spade is

P(S′)	= 1 - 1 / 4
	= 3 / 4

Dependent Events: Two (or more) events are dependant if the probability of the second event depends on what happened in the first event.

Example: A bag has 3 red and 3 blue marbles. A marble is drawn and not replaced. What is the probability that the second marble is blue?
It depends on what the first marble was.

Independent Events: Two (or more) events are independent if the outcome of the second event is unaffected by the outcome of the first event.

Example: Given the same bag of marbles. A marble is drawn and replaced. What is the probability that the second marble is blue?
Since these events are independent, the sample space is unchanged. The probability remains the same:

P(B)	= 3 / 6
	= 1 / 2

Mutually Exclusive Events: Two (or more) events are mutually exclusive if they have no events in common. I.e. They have no intersection (meaning the circles don't overlap eachother).

Example:

Calculating Probabilities using “AND” and “OR”
Given 2 events A and B

P(A and B) = P(A) · P(B)

If the events A and B are independent.

If A and B are dependant, then

P(A and B) = P(A) · P(B | A)

P(B | A) means “ the probability that B will occur given that A has already occurred.

Note: Tree diagrams are very helpful in solving these types of problems.

Examples: A bag contains 3 red & 3 blue marbles. Two marbles are drawn.

Independent Events:
Find the probability that the 2nd marble is blue if the first was red and had been replaced.

Dependant Events:
What if the first marble was not replaced?

P(A or B) = P(A) + P(B) - P(A and B)

When the events A and B are mutually exclusive P(A and B) always equals to zero. (There is no intersection/nothing in common) which simplifies to the formula above to

P(A or B) = P(A) + P(B)

Example: Flip a coin twice. What is the probability of getting two heads or tails?

So that’s all that happened in class. And to everyone in our class, the online quiz on quizstar is not going to be worth marks. Mr.K told me after class and decided not to, as the highest mark was only 50%. Yeah, I am just as surprised as everyone. If you’re curious to see what the answers are (with my explanation), you can click here.

Okay guys. It’s time for a wrap. The next scribe will be abr13l. No drum rolls, no flashy text, no cheerleading, I try to be as direct as possible ;). Homework is on page 380 in our blue textbook, questions 14-20. See you Monday guys.

blogging on blogging

2006-05-05T08:23:00.000-05:00

Wow i almost forgot about this blogging on blogging thing. Well I am finding this unit neutral for me. I understand the binomial theorum and how to find the such and such term. However the thing that is bugging me is when you are asked to find the number of ways to do this, using the choose or pick formula. I dont know when to use which formula. Overall, this unit is very interesting and it got me playing poker. hahaha!

blog post

2006-05-04T23:13:00.000-05:00

at last! i remember my username & password after inputting a lot of possible usernames and passwords that i'm using at blogger. i thought i can't log in anymore on my account. well, this unit seems easy for me at first because i had studied pascal's triangle and binomial theorem 2 years ago. but that term thing confuse me still sometimes, whether there is a 0th term or non. what confuses me also is that permutation and combination thing. there are times when i'm using "choose formula" in a problem where "pick formula" should be applied. i sometimes think that some problems are really impossible to solve but the group discussion helps me cause everyone is explaining their part leading to one answer favoured by the majority in our group. it also helps me to have an idea (and remain that in my mind cause we brainstorm for every question) how to solve a similar problem. hope to get a good mark for our test in this unit and hope i can still take the pre-test that i miss this afternoon.

Marcquins blogging on blogging

2006-05-04T22:37:00.000-05:00

so, I thought counting would be as easy as 1,2,3 but I was proved otherwise. I am catching on to the whole counting thing, nothing a little more homework can fix. So? The whole picking and choosing thing I still don't get when to use which one. Like how do I know if it is seven choose five or seven pick five? I'll be fine though, I will manage..........for now.... muahahaha!

Blog, Blog, Blog

2006-05-04T22:32:00.001-05:00

I don't quite think I'm fully prepared for the test. Wait let me rephrase that; I KNOW I'm not ready for the test. I've missed so many classes that I don't understand much of it, which I suppose is my own fault. I'm barely getting by, which might even be an exaggeration, but I do have my own sense of logic to figure out a couple of things on my own. I only pray to the Lord that I'll pull through this test, or better yet, this whole class.

bloggin on bloggin

2006-05-04T22:28:00.000-05:00

I don't get the counting unit I'm unsure of which formula to use on which. I sort of get the binomial therom part where you have to find the term given but when your expanding it and it goes into that long equation that's weird. Oh and I''m gonna fail the test I'm pretty sure, depending on my math skills the odds are againts me I was gonna use some sort of probability equation to show how much the odds are againts me but I don't understand much of the probability unit as well. Good luck to everyone, no need to wish good luck to me I'm already gonna fail it hahahaha hopefully I can understand it before the test begins.

blogging on blogging.

2006-05-04T22:16:00.000-05:00

well lets start off with this one. i dont feel that confident about this topic. i believe that i've been slacking off again.. and that is effecting me. so im going to start slacking. well to the i need help part of this. i think i have most of this stuff in hand, its just when it comes to dealing with the actual problems and using the equations i get mixed up on which to use.

blogging on blogging

2006-05-04T22:12:00.000-05:00

Well, i haven't been attending class regularly so i'm totally lost for the counting and probability. It's hard to grasp the concepts and it does get confusing at times. I get confused with the wording of the problems but that's my own fault since i kept missing the important stuff we learn in class.

bloggin on bloggin

2006-05-04T20:40:00.000-05:00

i kno this is too late to do but i just have to do it for the mark. for this unit i dont think im having problems but its just something that i can't understand. if i study or do homework its just the same hard and complicating. thats why it took me so long to do this blog cuz i didn't kno wut i was havin problems with. and i still don't kno.Counting part is alright, permutations same thing, pascal's triangle i will give it a 5. i really dont kno wut study for this unit. i thing this might be the unit i've studied for in this semester.
anyways i hope tomorrow test goes easy on us or special on me.
well good luck on everyone, i kno i need it.

counting blogging on blogging

2006-05-04T17:14:00.000-05:00

Counting is over...*sigh. I think this unit could be the hardest one yet for me. Today was a one period class and all we did was our pre test. Our group did good with the multiple choice questions but we didn't get any of the long answers at all. We made small minor errors that really made a big difference in our answers.

I definitely understand the concepts but I just don't know when or where to use it. Especially the questions that have one or more restrictions. When I see questions like those I don't know where to start. What I'm really confident about this unit though is the Binomial Theorem. I hope we get a lot of those in the test tomorrow.