<p>were there two different experimentals for math
my friend got another grid in
but i got a normal extra math section</p>
<p>what?!? it was definitely 45 9th graders, wasn’t it?</p>
<p>for the triangle one wouldnt you use triangle inequality theorem?</p>
<p>6+8 = 14</p>
<p>8-6 = 2 </p>
<p>2<x<14 right? Because its asking one possible value?</p>
<p>What was the perimeter of the triangle one where it have you 2(a+b)+c=38 and it have you two more that equaled 40 and 42?</p>
<p>no 45 was the number of 9th graders with a locker in B, the question wanted total students in hall b</p>
<p>exactly it asked how many students were in hall b and stated that 20% were ninth graders. Therefore, if 20% were ninth graders it can’t be 225 it has to be 290</p>
<p>@Phasmatis: The question indeed asked the number in the row for the Hall B not the number of freshman column…</p>
<hr>
<p>i thought i got all right but apparently -2
■■■</p>
<p>^I feel the same as you teddybear–exactly the same.</p>
<p>the one with slope and the angle thirty was x/rad3x. It cant be just x/rad3. lets say x equals 2, the the slope would be 2/2rad3 not 2/rad3. i’m pretty sure it didnt state that x=1</p>
<p>What was the answer for the shaded/unshaded circle where you had to find x, the number of degrees in the unshaded part of the circle. Or was that experimental?</p>
<p>^^ there wasnt an answer x/rad3x there was x*rad3x</p>
<p>i got 5 for the toothpick but im second guessing myself because it was such an easy problem so late in the section</p>
<p>the triangle one:
Given triangle is 8-12-10 total perimeter is 30
smaller triangle is 4-6 and x so total is 10+x</p>
<p>30/10 = (10+x)/(x)</p>
<p>cross multiply and everything and you get x=5
5+10=15 is what I put</p>
<p>it is 290.</p>
<p>45+70+75+100</p>
<p>it said that 20% of the NINETH GRADERS were hall B.
and it was asking the TOTAL number of students assigned to hall B.</p>
<p>the table across the row Hall b was 45,70,75,100</p>
<p>can’t you think of it this way?</p>
<p>you need to throw three dice, to get 18 as sum, you need three 6s. The probability of getting a 6 on first throw is 1/6, second throw 1/6, third throw 1/6. Thus 1/216</p>
<p>To get 17 as sum you need two 6s and one 5, probability of getting a 6 or 5 first throw is 1/6, then 1/6, then 1/6. Thus, 1/216</p>
<p>You add both probabilities because you add probabilities, usually, when it says OR thus giving us 1/108</p>
<p>This is my approach. However i am kind of convinced about 1/54.</p>
<p>Why is this approach wrong.</p>
<p>1000% sure it is not 11-3x-y
maybe 11-4x-y but there is for sure no 3</p>
<p>@spt.leonidas…I actually don’t remember what the 290 was and what the 225 was for. i know that 45 was freshman in B and that is not the answer cuz they wanted the total. I forgot what I put down as the total. was 225 the sum of freshman or sum of hall b? what about 290?</p>
<p>@oblivi0n i did/thought of it exactly that way! but after looking at this forum, i’m thinking otherwise.</p>
<p>@ ski-kid, it was 5, i hand-counted like an idiot to make sure :). I think it was only on the second page though</p>