<p>Def890, was it asecx? I don’t remember any problem that had 9secx as a choice…</p>
<p>yea</p>
<p>btw it’s a*sec(x) not 9 :)</p>
<p>did anyone get:
A, 2x-y=0 for parallel lines
10*tan66, C</p>
<p>post your #1-20 harder questions/answers because missing those are a real killer :</p>
<p>im sorry, yes it was a sec x…haha sorry</p>
<p>Yeah, I got the above answers. I’m like devastated over my test right now so I don’t want to remember any more questions. What’s the best score prognosis for like 10 wrong? =/</p>
<p>aznosamaboy, i got those two answers too</p>
<p>would 10 wrong be a raw score of 38? thats probably around a 750 or so</p>
<p>Yeah, that’s what I thought and according to Sparknotes. I am so wishing for a better curve.</p>
<p>waiting 2.5 weeks will suck</p>
<p>i omitted out least 10, and maybe got a few wrong. Do u guys think i will get at least a 700? thats waht i’m aiming for. :(</p>
<p>what was the answer to the standard deviation problem?</p>
<p>i put the answer that had all the graphs equal to each other, because standard deviation is how much the data deviates from the average, right?</p>
<p>No. the standard deviation one was the graph with one bar. If all the data is the same, then there is no deviation.</p>
<p>yeah, the answer was the one bar…it was letter choice A</p>
<p>wasn’t the data all the same in the one with all the graphs equal to each other. i didn’t look clearly at what each graph represented, like the units or anything, just the shape. if one of the graphs is high, and the other two are 0, then would that mean the standard deviation is higher?</p>
<p>The one with the same bar lengths had a different number of students for each bar, which means there is standard deviation.</p>
<p>what was the answer to that question with the prime numbers q3t5w7</p>
<p>was it q9t15w21? or something like that?</p>
<p>damn! i didn’t look at the number of students for each bar! i only saw the fact that they were equal legnths and assumed they were equal. crap.</p>
<p>the question where it asked u to find the midpoint coordinates and they gave u two points, (x,y,z) (x2, y2, z2). do u jus average thema ll together?</p>
<p>thanks.</p>
<p>Yeah, I’m pretty sure that one was (5,0,6)
You get it the same way as normal midpoint, just adding z1+z2/2 to get the 3rd point.</p>
<p>Yeah, I remember the answer being like A, which was like (5,0,6) or something?</p>