<p>Shouldn’t the A+B+C question be 8? The given digits add up to 22, so the only way for the final number to be a multiple of 3 is to make it equal one. 22+8=30, which is mult. 3.</p>
<p>Range = 2 x amplitude.
8.6</p>
<p>What was the whole sin problem? Can someone type it up? I see answers -.64 and I don’t remember what i put.</p>
<p>cos^2x + 1 = .64</p>
<p>Actually it was 1 - (cos(stuff))^2 = .64, so what does (sin(stuff))^2 equal?</p>
<p>Answer is .64</p>
<p>Oh lol because you just substitute 1- (cosx)^2 for (sinx)^2 ok thank goodness</p>
<p>Is .64 or -.64 the right answer for the sin problem? Someone clears this for me.
@JD: You mistook the question. It said that long number 4,ABC,xxx,xxx is the product of 6 consecutive number multiple of 3. So it should be like 24 x 27 x 30 x 33 x 36 x 39 = 4,ABC,XXX,XXX.</p>
<p>are u guys saying -5 as 5 blank or 5 wrong</p>
<p>Wish everyone good luck and the best scores!!
Really glad I saw your answers, makes me feel a lot more confident about my score. I think I might have got the ambiguous case triangle wrong it wasn’t 131 degrees was it?</p>
<p>Also is there a general agreement about the percentile question? Was it in between or not possible I put inbetween</p>
<p>Finally I left the probability question at the end empty. How would that have been answered where 1 event can occur but not both?</p>
<p>Thanks everyone</p>
<p>@SATnotACT: I don’t actually remember the equation. But I remember that .64 ( or -.64) in the question was also the answer.</p>
<p>darksouls you find the likelihood of both by mulitplying both together, find respective probabilities and then sum them to get .76</p>
<p>how would sin squared x equal a negative. it will be positive.</p>
<p>.64</p>
<p>@HBOUND
Thanks but I omitted that question. I hope that question isnt the difference between a 790 and an 800 lol</p>
<p>@Darksouls . The probability question at the end was .52. You have to subtract the probability of getting both and the probability of get neither from 1 so the only value left is the probaility of getting only one or the other. So 1-((.6x.4)+(1-.6)(1-.4)) and you get .52.</p>
<p>mbomb99, how can the probability of getting at least one be less than .6, the probability of getting the textbooks? </p>
<p>The answer is .76
It helps to use a venn diagram to visualize</p>
<p>Mbomb, it is at least 1 object. The easiest thing would be p(not a and c) which would give you .76</p>
<p>Yeah but you have to take away the probability of getting both then the probability of getting neither and THE ONLY thing thats left is probability of either or. Theres nothing else. So it is .52, also this was a question on barrons about football nfc and afc.</p>
<p>The question asked what is the probability of getting at least one of the two, not either one or the other. So you don’t subtract both.</p>
<p>Thats how the question was done in barrons. What did you get for the that ask"7^6=10R-S " i believe and s is less than 10 and is an integee</p>
<p>^ Is 9. Btw, why did some of you guys get 7.8% on problem 50(the one about population increase)? I got 7.78% I think. I did (1+0.003)^25=1.07776… which is rounded to 107.78%, an increase of 7.78%…I remember putting B…</p>