SAT May 2009 Math

<p>would 3.38 be wrong? i didnt think of putting it into fraction form…</p>

<p>ceballos, I and II were correct, i think…</p>

<p>what was the 28/7 question?
yeah that list above should be right</p>

<p>I, II, and III</p>

<p>was I and II only.</p>

<p>oh yeah, i put all three i think .</p>

<p>I and II…anyone?</p>

<p>a is divisible by b. b is divisible by c…</p>

<p>I
II
III</p>

<p>"What was the one with I II and III? I forget what the question was, but II was “has at least three factors”</p>

<p>I said I and II, but I couldn’t remember if “1” counts as a factor…</p>

<p>dangit i got that one wrong then… i put all three</p>

<p>this tin and copper question is really annoying me; and what question is 28/7; i dont remember that one</p>

<p>mode < mean < median?</p>

<p>The venn chart “exactly 2” was just add all the ones where the charts intersected?</p>

<p>yeah it is 10
whats the I II II answer guys?</p>

<p>It’s definitely I and II only</p>

<p>3x6 = 8</p>

<p>lowest common multiple = 12</p>

<p>Use 4 2 and 1 for example, it works for I and II but 4 != 2 x 1 so III is incorrect?</p>

<p>for the I II III i plugged in 5,15,45 and 3,9,27 and all 3 worked so i put all three. but i guess i’m wrong? 2 wrong so far… ugh</p>

<p>I put I and II as well. The one that was crossed out III (a=bc).</p>

<ol>
<li>A is divisible by C</li>
<li>A has at least three factors</li>
<li>a=bc</li>
</ol>

<p>a definitely does not equal bc.</p>

<p>it said odd integers… for I II III problem, didn’t it?</p>

<p>Still, it doesn’t matter. Use 9, 3, 1. 9/3 = a/b, works. 3/1 = b/c, works. 3*1 != 9.</p>

<p>For the ‘a’ is divisible by ‘b’, ‘b’ is divisible by ‘c’, I think it was I only. Note that a number is divisible by itself, so what if a, b, and c were all, for example, 7?</p>

<p>I believe it said that they had to be different.</p>