<p>is there a consolidated list? if so, what page number, please help me out</p>
<p>Though I answered 20, I found out a few hours later that the answer is indeed 24.</p>
<ol>
<li> 12354</li>
<li> 12435</li>
<li> 12453</li>
<li> 12534</li>
<li> 12543</li>
<li> 13254</li>
<li> 13245</li>
<li> 13452</li>
<li> 13425</li>
<li> 13542</li>
<li> 13524</li>
<li> 14235</li>
<li> 14253</li>
<li> 14325</li>
<li> 14352</li>
<li> 14523</li>
<li> 14532</li>
<li> 15243</li>
<li> 15234</li>
<li> 15342</li>
<li> 15324</li>
<li> 15432</li>
<li> 15423</li>
</ol>
<p>^is that the answer for the marching bands q.? I put 24.</p>
<p>And pls. consolidated list pg. no. pls.</p>
<p>nvm found it. pg.55</p>
<p>Fill-Ins:
1.2 = radius
m = 2
1080 average</p>
<p>^what are those 3 q’s? Can anyone refresh my memory?</p>
<p>what was the marchingband questions?</p>
<p>also, this seems ridiculous, but does anyone remember what number grid in the obtuse triangle one was?</p>
<p>1.2 eng quart = 1 brit quart</p>
<p>^didn’t this questions ask how many UK quarts one US quart was? I got 5/6 I think.</p>
<p>@ viggyram</p>
<p>the 1.2 radius was a fill in for one big circle and a little circle with a shared line across it. </p>
<p>m=2 was the second had to do with factoring out. How i did it i got -5 and 2 or something. but 2 was the answer because negative can’t be gridded. </p>
<p>No sure how the 1080 question went.</p>
<p>^I don’t remember any of those…I don’t know why.</p>
<p>anyone remember how the (-3,3) ax^3 - 9 question went?</p>
<p>What do you guys think the curve will be? Specifically, -2 (1 MC and 1 FR)?!</p>
<p>This thread is making me so nervous 
. Before coming on this thread I was confident I had a 800, yesterday I realized that I probably missed one, now I’m hoping that I only missed 2.</p>
<p>consolidated list please?</p>
<p>I wrote this out earlier, but post #935 seems to have it covered.</p>
<p>I didn’t take this test, but from what reasoning I’ve done, I support the 1/54 answer for the dice question. </p>
<p>We can all agree that there are 216 possible outcomes for the dice. Of those 216 possible outcomes, four yield a sum of 17 or 18. If we visialize the three dice being rolled simultaneously, and pretend that they line up evenly from left to right, then they would look like 111, 112, 113, etc. The four possible results that yield 17 or 18 are 666, 665, 656, and 566, which of course yields a probability of 4/216 or 1/54. The reason 1/108 doesn’t make sense to me is that it implies there are only two possible ways to get 17 or 18. We can all agree there is only one way to get 18, so for 1/108 to be true there must be only one way to obtain 17. This is incorrect, because of the 216 possibilities, three yield 17. Not one.</p>
<p>Another way to visualize this problem (and disprove the 1/108 choice) is to imagine the data of thousands of three-dice-roll simulations arranged into a distribution chart according to the sum of the die. Each end would be the extreme values 3 and 18, which would have the lowest number of occurences. Conversely, the sums toward the center (like ten and eleven) would have the highest number of occurences, as there are many possible ways to obtain these sums. Essentially, this distribution would resemble a bell curve: as you approach the the middle sums from the outside sums, the number of occurrences (and therefore the probability) of the sum increases.</p>
<p>Following this train of logic, the probability of obtaining a sum of 17 must be greater than the probability of obtaining a sum of 18, by the very nature of the bell curve. We know that the probability of a sum of 18 is 1/216, so for 1/108 to be the answer, the probability of a 17 sum must also be 1/216. However, this cannot possibly be the case, as 17 is closer to the center of the curve than 18 and therefore must have a higher probability of occurring.</p>
<p>The reason (1/6)<em>(1/6)</em>(2/6) doesn’t work:</p>
<p>Let’s pretend the dice are numbered 1-3, and every time you roll them, they line up 1-3 from left to right. Using (1/6)<em>(1/6)</em>(2/6) (each probability corresponding to dice 1, 2, and 3 respectively) is insufficient, because it means only die number three could possibly roll a 5. It fails to take into account that die number one or two could also roll a 5.</p>
<p>Anyone remember getting a .763 or .764 for the grid-ins? I’m kind of worried because I don’t ever remember getting an irrational number for a grid-in answer.</p>
<p>^that definitely wasn’t an answer.</p>
<p>Did anyone else get these questions? I notice no one has mentioned these at all, I think they are an experimental section:</p>
<ol>
<li>What is the perimeter of a triangle with sides a,b, and c if:</li>
</ol>
<p>2(a+b) + c = 38
2(a+c) + b = 40
2(b+c) + a = 42</p>
<ol>
<li>What is the number of combinations of a 3-digit number that is less than 600 that can be made from the numbers 3,5,7, and 9?</li>
</ol>
<p>^ flyingfeathers, those questions weren’t on my test, so they were probably experimental. but if you’re looking for the answers, for number 1 you add the 3 equations and divide by 3. for number 2 i suppose you could just count.</p>
<p>So I think I got the dice one wrong.
What score would -1 be?</p>
<p>It’s ok if I didn’t simplify a fraction in one of the grid-in questions, right?</p>