<p>It didnt say 3 numbers lol. The number was never specified.</p>
<p>Was the answer to that tic-tac-toe question with the X’s 6?</p>
<p>haha ;)</p>
<p>@presidont: I didn’t get that question…so i’m guessing it was on the experimental section</p>
<p>Presidont you are correct. It was experimental though I think.</p>
<p>Oh, the average problem. I think it told us that the average came out to 15, and then if you added 20 to one of the values it came out to 19 as the average.</p>
<p>s = sum of the values, n = number of terms</p>
<p>Eq. 1: s / n = 15</p>
<p>So adding 20 to one term basically just means increasing the sum to 20.</p>
<p>s + 20 / n = 19
19n = s + 20 (multiply both sides by 15 to eliminate the denominator)</p>
<p>Isolate s:</p>
<p>Eq. 2: s = 19n - 20</p>
<p>Plug that back into Eq. 1: </p>
<p>19n - 20 / n = 15
15n = 19n - 20 (multiply both by n to get rid of the denominator)
4n = 20; n = 5</p>
<p>So now you know there are 5 terms. Plug 5 back into n for Eq. 1 and you can easily find the original sum, which was 75.</p>
<p>Did u guys have the american SAT? I took mine in ontario, canada, so is that the same/different? Because I don’t know what any of you are talking about…so I’m assuming I had a totally different math SAT?</p>
<p>yudawg, I got the triangular prism problem, and I had CR experimental. The answer is 26.</p>
<p>I put 109 for the one where you kept having to multiply the preceding term by 2; I was kind of iffy on that one. :/</p>
<p>@pchs10: I think it might have been experimental because I had two math sections with grid ins. </p>
<p>Also because there are apparently two completely different SAT tests do you guys think that are two different distinct experimental math sections?</p>
<p>For the average one:</p>
<p>It did not specify how many numbers there were. This was how I approached the problem:</p>
<p>Let s be the sum of all the values, and let n me the number of values. Therefore</p>
<p>average = s/n</p>
<p>Now, since you are adding 20 to one number, and the new average is 19, you have:</p>
<p>19 = (s+20)/n
19 = s/n + 20/n</p>
<p>Replace s/n with the given average.</p>
<p>19 = 15 + 20/n
4 = 20/n
n = 20/4
n= 5</p>
<p>The average of five numbers is 15; we can say that the numbers are all 15. So that must mean that s = 5*15 = 75.</p>
<p>EDIT: Chasel beat me to it! :P</p>
<p>Yeah it seemed too easy for it to be the last question lol. You put D right?</p>
<p>Well, I had a math experimental, but some ppl are saying that the math experimental was the one with the “horse” question…which I didn’t have. So I’m assuming the test I took was the international SAT, and thus had completely different math sections>?</p>
<p>@yudawg: I had a math experimental section but I had only 1 grid-in section.</p>
<p>What was the original question. My ■■■■■■■■ brain can’t read right</p>
<p>What about the calculus class question? 1998, 1999, 2000</p>
<p>1999 had 5 more students than 1998. 2000 had 20% more students than 1999. The only way to get an integer result is for 1998 to be 10.</p>
<p>The answer is 10, or A.</p>
<p>The square question could easily be one. Why?Simple, the smaller square had a side of sqr root of two. The larger square had an area of three and thereby, a side of 1.5 the diameter of the circle had to be more than half the side of the smaller square but less than half the side of the bigger square. Half of the sqrt of two is .70 something and 1 is greater than .70 but less than 1.5.
therefore one could be a possible answer.</p>
<p>no, because you have to count in the diagnols. Trust me, it HAS to be 1<x<1.5 .</p>
<p>anyone know the answer to the last question on the 20 minute section? </p>
<p>f(0)=?</p>