### 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, -x

^{2}, x

^{3}, -x

^{4},

__x__,

^{5}__-x__,

^{6}__x__

^{7}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..

10!/5!5!

the 5! represents the correct ones and the other 5! represents the wrong ones. you would then get..(C)

^{5}(W)

^{5}

(0.3)

^{5}(0.7)

^{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

=0.73

= 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 (:

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