<p>somebody please answer why the prob question would be 1/6 and not 16/81=(4/9)^2</p>
<p>i know the answer was cos(x)=-4/5! that is the consensus! but i dont know if i put that down. all i remember was taking the x/r, so if the 4 was there then i MIGHT’VE gotten it right. i honestly dont even know how to do the problem.</p>
<p>awkturtle, you probably got a 790 if you omitted 2 and got 3 wrong. i dont feel too good about my score >.<</p>
<p>He meant to say 135 both times, right?</p>
<p>yay! i hope i answered that one because i don’t remember haha.</p>
<p>if the log one is the one i’m thinking of i think i got 4950… :o</p>
<p>does anyone remember the answer to the 2^2x=k one?
was it like x= log k-log 4?</p>
<p>i would LOVE a 790. im sure you did ok. the curves is usually pretty fair right? i got a 650 last year because i didn’t know you could omit questions! lesson learned: SKIP questions instead of randomly bubbling in c. haha.</p>
<p>does anyone remeber what the three answers to the dog problem where
- x>6 formula 2 is lower
- ??
- g(10)= some number other then 55</p>
<p>and log questions was log(k)-log(4) i think? they didn’t put base in the answer so I assumed 2 because if it was ln i don’t think any worked</p>
<p>IF its without replacement then it becomes</p>
<p>(4/9)(3/8) = 1/6</p>
<p>but they never mentioned taking the number out once you choose it the first time…</p>
<p>dgor91, the chance of picking an even number the first time is 4/9. the problem says that the second time a number is picked, it can’t be the same even number as the first one. so the chance of picking an even number the second time is 3/8.
(4/9)(3/8)=1/6</p>
<p>I think we established that the P(even numbers) is ambiguous, no one remembers the exact wording of the question. It’s either 16/81 or 1/6.</p>
<p>ya i do NOT remember them saying explicitly “the second time a number is picked, it can’t be the same even number as the first one” or select “two DIFFERENT numbers”…</p>
<p>i think the replacement/not replacement thing is slightly confusing. but … think about it like this. if they told you that the second number cannot be the same as the first even number, then it will definitely have a different probably than if the problem just asked you: what is the chance of picking any even number the first and second time?</p>
<p>they said two DIFFIERENT numbers. 99.9% sure</p>
<p>they did say that the second number had to be different from the first, this i remember CLEARLY</p>
<p>i remember that it definitely didn’t say that the number was replaced but i don’t remember it explicitly stating that you would pick a different number…</p>
<p>“two different numbers” is the exact wording they used.</p>
<p>So dgor91, the answer is 1/6 because the first number chosen is essentially “removed” from the set.</p>
<p>Has there been a consensus on the polar coordinants problem? Was the answer E?
Also can someone remind me what the prime number question was asking?</p>
<p>yeah it was e</p>
<p>If r and s are different prime numbers, what COULD produce another prime number.</p>
<p>i remember 2 answers were:
none.
r + s + 1</p>
<p>I guessed none, but the answer is r+s+1.
3+5+1=9 (not prime), but 5+7+1=13 (prime)</p>
<p>So r+s+1 is correct because it could produce a prime number.</p>
<p>remember that question where x / (2y - x) = …, so what is y / (2x - y)? It’s the only one I omitted because I figured the algebra would take too long. What did you guys get?</p>