January 15, 2007

scribe post; geometric sequences.

today we started off with mr. k giving us questions on the board.
find the next three numbers..
1, 2, 4, 8, 16, 32
how you would get the next set of numbers was multiplying the number by two
OR 1, 2, 4, 7, 11, 16
how you would get this one was to find the pattern in the sequence. so you would have to find the difference between the numbers. for example: 2 and 1, the difference is 1. and the difference between 4 and 2 is 2. so you would get the pattern. you know that the number is increasing sequentially. so the next number must be a 7 - since the next number is a 7, the difference between 7 and 4 is three. and so on..

6, 11, 16, 21, 26, 31
for this one, the difference between the numbers is 5 and it's always constant.

3, 6, 12, 24, 48, 96
this one is basically like the first one, you multiply the numbers by two.

1, 1, 2, 3, 5, 8, 13, 21
for this one.. you find a pattern by adding the numbers.. for instance: 1 plus 1 is 2. then 2 plus 3 is 5. and 3 plus 5 is 8. so you get the point..

1, 3, 6, 10, 15, 21, 28
this one is basically like the first one.. but the second answer.. but this one doesn't start at 1. it starts at two - meaning.. the difference between 3 and 1 is 2. 6 and 3 is 3. and it's going up sequentially, and so on..

x, -x2, x3, -x4, x5, -x6, x7
for this one, you would multiply by -(-x)

3, 4, 7, 11, 18, 29, 47
this one, you add the numbers together. 3 and 4 is 7. 4 and 7 is 11. 7 and 11 is 18. etc etc..

77, 49, 36, 18, 8, 8, 8
this one was kinda tricky. you would split the numbers - so you would get 7 and 7. you would then multiply them together and you would get 49. 4 times 9 is 36. 3 times 6 is 18, and soo on..

in the afternoon class, we had a pretest on probability. and here are the questions and solutions to them.
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

first of all, YOU have to go. so you would get 1 x 11C2 because you're part of that twelve so there are eleven people remaining to choose from - and you're going to choose two. and the denominator would be 12C3 since there are twelve people and you have to choose three. so that would look like..
1 x 11C2 ALL OVER 12C3
and that would all work out to 1/4, which would be a.

2. rex is playing a guessing game. the probability he will guess each question correct is 0.3. what is the probability he will guess exactly 5 out of 10 questions correctly?
a) 0.15 b) 0.10 c) 0.29 d) 0.80

so the P(correct)= 0.3 and the P(wrong)= 0.7
since there can be different ways of getting the questions wrong or right, but you still have to get five correct. so you would get..
the 5! represents the correct ones and the other 5! represents the wrong ones. you would then get..(C)5(W)5
and when you work that all out, you would get 0.10 and that's the answer b.

3. you choose four digits from the numbers 0-9 (ten digits in total), with no repeated digits. to win a prize, all four of your numbers need to math the four selected by a computer. the probability that you will win a prize is:
P(all four digits)= 4C4 ALL OVER 10C4
and when you work that all out, you would get 1/210 (and that's the answer)

4. the serial nuber of a $10 bill contains 8 digits. if your $10 bill contains the digit 7 at least once, you will win a prize. what is the probability that your $10 bill will win?
first find the probability that you wouldn't get any 7's. so, the probability that you would get a 7 is 1/10 and the probability that you wouldn't get a 7 is 9/10. soo..
P(not seven)= (0.9)8
since you now have the probability of not getting any 7's, you can now get the probability of getting at least one 7.
P(7 at least once)= 1-(0.9)8
so, you would get 0.57 (and that's your answer)

5. jeff takes a lunch to school on two days, and on the other three days he buys it. if he takes lunch - he is late for his period 4 class 15% of the time, but if he buys lunch - he is late 35% of the time.
a) what is the probability that jeff arrives on time?

b( if jeff was late, what is the probability that he took his lunch to school?
first make a tree diagram

P(arrives ontime)= P(takes&ontime) + P(buys&ontime)
= (0.4)(0.85) + (0.6)(0.65)
= 0.34 + 0.39
= 73%

P(late|took)= P(takes&late)/P(takes&late) + P(buys&late)
= 0.06/0.06 + 0.21
= 0.06/0.27
= 0.2222
= 22%

oh yeah, the test is now on wednesday.
umm.. the next scriber is tennyson.
gooday&goodnight (:

No comments:

Post a Comment