Need shortcuts/ help Math 2

Test Preparation SAT Subject Test Preparation
1

<p>23) If a fair coin is flipped three times, what is the probability that the results will be tails exatcly twice?
(any shorcuts that would not waste time finding all the possible outcomes?)</p>

<p>50) Seven blue marbles and six red marbles are held in a single container. Marbles are randomly selected one at a time and not returned. If the first 2 marbles selected are both blue, what is the probability that at least two red marbles will be chosen in the next three selections?</p>

<p>43) The system of equations given by
2x + 3y = 7
10x + cy = 3
has the solution for all values of c except
-15, -3, 3, 10, 15</p>

<p>44) If f(x,y) = xy/3 for all x,y f(a,b) = 15, f(b,c)= 20 and f(a,c) = 10, which of the following could be the product of a, b, c?</p>

<p>45) if x>0 and y>1, then log(y)/log(x^2) =
I. log(y^2)/log(x)
II. log(sq root of y)/log(x)
III. log(y/2)/log(x)</p>

2

<p>dude PR?</p>

<p>do u understand number 40 in practice test A</p>

3

<p>23) (3 choose 2)(0.5)^3</p>

4

<p>23) 3C2 / 8 8 b/c 2<em>2</em>2 3C2 b/c 3 flips, choose 2 tails</p>

<p>50) start by subtracting the 2 blues...so you have 5 B, 6R</p>

<p>[ (5C2)(6C1) + (5C3)(6C) ] / 11C3</p>

<p>43) multiply the top row so it has one term equal (multiply by 5 so you have 10x in both rows) it can't be 15 b/c 10x + 15y = 35 cannot be solved if 10x + 15y also = 3</p>

5

<p>to understand 40 you would need a graphing calc, thats how i can explain</p>

6

<p>how could we explain it if we don't know the answers?</p>

7

<p>44) ab/3 = 15 bc/3 = 20 ac/3 = 10 solve for each, then multiply</p>

<p>45) pick some values for x and y, then plug in. i'm too lazy to solve that one</p>

8

<p>what is C? is it nCr?</p>

9

<p>what does three choose two mean?</p>

10

<p>fine i will give the answers</p>

<p>23) 3/8</p>

<p>50)19/33</p>

<p>43)15</p>

<p>44)284.60</p>

<p>45) only I</p>

11

<p>3 choose 2 is nCr I don't know how else to explain it. It's the counting principle - how many ways can you choose two out of 3? (choose 2 tails out of 3 flips)</p>

12

<p>I did 3 nCr 2 on my calc and its giving me 3</p>

13

<p>how do you do 23? i mean with only 3 coins it is easy to picture out (TTH, THT, HTT)...but how would you do it with more coins? Is there like a formula for it?</p>

14

<p>3C2 is 3.</p>

<p>for 23...<br>
(#flips CHOOSE #you want heads/tails) use calc nCr<br>
/
(2^#flips)</p>

15

<p>That is what Im trying to ask</p>

16

<p>to visualize 23, there are 8 ways it could happen, three favorable</p>

<p>HTT THT TTH ttt hhh thh hth hht</p>

17

<p>^sophia can you explain how you did number 50 im kinda confused with that one... thanks!</p>

18

<p>(n choose k)<em>(p^k)</em>(1-p)^(n-k)</p>

<p>n = number of trials
k = number of times a certain even happens
p = probability of that event happening</p>

<p>on a graphing calc...
2nd VARS(distr)
binompdf(n,p,k)</p>

19

<p>the "formula" in post #16 is for anything with the coins</p>

<p>for #50, "Seven blue marbles and six red marbles are held in a single container. Marbles are randomly selected one at a time and not returned. If the first 2 marbles selected are both blue, what is the probability that at least two red marbles will be chosen in the next three selections?"</p>

<p>The fact that the first 2 are blue has nothing to do with the answer, other than to confuse you...you just subtract 2 blue from the original data. You are left with 5 blue and 6 red.</p>

<p>First, you have to figure out the number of ways you could pick 2 or more reds (and since you are picking 3, that means 2 or 3 reds). Then you add those ways up, and divide by the total number of outcomes.</p>

<p>For 2 reds: (6 red C 2) (5 blue C 1)
For 3 reds: (6 red C 3) (5 blue C 0)</p>

<p>Total outcomes: (11 total C 3)</p>

20

<p>^ah much clearer now thanks so much!</p>