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Tuesday, June 13, 2006

Scribe guY

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

OR

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 7th 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

OR

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:

70

*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 =)




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1 Comments:

At 6/15/2006 5:08 PM, Blogger Hobo said...

That's so long....

 

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