<p>Uhhh correct me if I’m wrong but I don’t think (-3,3) was the answer to the ax^2 - 9 question. (-3,3) was given and it asked something else.</p>
<p>@jlewis</p>
<p>AHH thank you! (: i guessed correctly then. i was worried that there was a 1/72 for B xD</p>
<p>jlewis, i think the answer was 4/3 or something like that</p>
<p>^ ah, yeah that’s what I remember putting</p>
<p>Anyone remember the grid-in question that involved two different functions and you had to find out the least value of x for which the function was greater than the other function? I think the answer was 8.</p>
<p>and two times a number plus 8? idr, but it added up to 23 i think it was 18 and 5</p>
<p>ohhh mannn apparently i misread the question.</p>
<p>Anyways, what is “27 people for survey” question? I don’t remember, oh and 3a-20</p>
<p>@jollybjolly: I put 8 for that.</p>
<p>So here’s my logic for the dice question.
To roll an 18, the dice must be 6,6,6. This is 1/6<em>1/6</em>1/6
To roll a 17, the dice must be ONE 5, and TWO 6. To roll a 5, there’s a 1/6 chance. To roll a 6, there’s again 1/6. Thus: 1/6<em>1/6</em>1/6</p>
<p>Add both of those together, 1/216 + 1/216 = 2/216 = 1/108</p>
<p>^ The problem with this argument is that you’re saying that only one die is allowed to roll a 5.</p>
<p>^^ Order matters for this problem. You have to realize this. The PROBABILITY of getting a 17 is 3/216 and an 18 is 1/216. It’s 1/54.</p>
<p>People, just accept the fact and get it over with.</p>
<p>The dice question seems rather ridiculous, but here I am to weigh in, even though I never took this test:</p>
<p>Let’s forget about 3 dice. Let’s pretend we only have 2. And we’re trying to figure out how many ways you can get a sum of 11 or 12. Now, I personally like to make a table, with one side going across horizontally as the value for die 1 and one side going down vertically for die 2. Then I fill in my chart adding the value from the left with the value on top. You would get 11 twice and 12 once. If you made boxes for your table to be extra neat, you would have 36 boxes (the area of the rectangular table). So, that would be a 3/36 probability, or 1/12.</p>
<p>Applying the same logic to this particular question, you would have a cube. Of course, you can’t possibly have enough time to draw a cube with all the dice rolls listed in mid-air. But, the same concept as 2 dice applies to 3 dice. For 2 dice to equal 12, you’d have that situation occur once in the bottom right corner with 6+6. The same would occur with 18. It would occur once in the bottom right hand corner with the ordered pair being (6,6,6). Now, how many points in this cube would have a 5, a 6, and a 6? You could have (5,6,6), (6,5,6), and (6,6,5). That’s 3 possibilities. So, your probability would be the number of possibilities over the volume of the 6-by-6-by-6 cube. 4/216=2/108=1/54.</p>
<p>My 2 cents, of course. This is just a graphical/table way of thinking of it (as opposed to that weird combination/permutation/order matters/order doesn’t matter s**t).</p>
<p>Was the section where you have to find the number of hours or so that the student studied experimental?</p>
<p>The dice problem screwed people who studied, and people who didn’t. The only people it didn’t screw are the ones who truly understand combinations.</p>
<p>two questions… for the TV question, was 27 C? i’m totally blanking… and also, i don’t remember the july/august question at ALL. this was definitely not experimental, right? does anyone remember more about the question?</p>
<p>oh and, does anyone remember the question about the grade levels… something about grades 1, 2, 3, 4, 5 vs 6 and 7?</p>
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<p>and people like me who just guessed and got it right :)</p>
<p>lol i just did 36(3)= 108 </p>
<p>and then there were 2 values (17,18) so i just divided by 2= 1/54</p>
<p>hope im right!</p>
<p>Ether, it doesnt matter? Cause either way, only ONE die has to be a 5. Doesnt matter which one, so its still a 1/6 chance of that happening.</p>
<p>Lol, I just guessed too, I put 1/54, sucks for the people who actually tried it</p>
<p>1/54 is the correct answer. 1/108 is a Joe blogger answer. End of debate for dice question.</p>