<p>I got 2 and 4 for the 2n is divisible by 8 too
N=20 40/8 =5
So 20 can’t be divisble by 8</p>
<p>I got A 5workers!! Hoping its correct.</p>
<p>I’m pretty sure that the average median problem of temp. is 93…</p>
<p>@journalismjunkie: idid the same as u 93 , but its wrong when i gave it another read the correct answer is 88. </p>
<p>& for the workers question it was 5.</p>
<p>till now i have 1w 1left 
750-760?</p>
<p>@kittie444: K - 4 is answer for (x-4)(x+4)</p>
<p>anyone remember what choice that was? is it “C”?</p>
<p>The parabola is (-3,0) for sure, the equation was x^2+kx+15 and the vertex was (4,-1) if you solve the equation its (x+5)(x+3) or x^2+8x+15 leaving the possible intercepts to be (-5,0) or (-3,0) and (-5,0) was not one of the option so its (-3,0) for sure.</p>
<p>@saswisher that is incorrect… as you said the equation is y=x^2+kx+15, with one of the points being (4, -1). if you substitute the points for the unknown variables in the given equation, you get
-1 = (4)^2+k(4)+15
-1 = 16 + 4k + 15
-1 = 31 + 4k
-32 = 4k
k = -8</p>
<p>then if you substitute -8 for k back in the original equation you get
y=x^2-8k+15, which, if you factor correctly, gives you (x-5)(x-3)=0, and from then you can clearly see that the x-intercepts are (5,0) and (3,0)</p>
<p>if you do not understand the above, you can look at it this way: since the point (4, -1) is the vertex, it’s x-value (4) is -b/2a
so
4 = -k/2
-k = 8
k = -8</p>
<p>and from then on you can solve the equation as i stated above.</p>
<p>the median was 88…
there was an even number of temperatures
like
73 75 80 (88) 90 93 95
and the median was 88 so it fits - its the middle number
and ^ agreed. got (5,0) as well.</p>
<p>How is the answer to the probability question 0.4?</p>
<p>Wasn’t it 2/5? What did you get Kyoto</p>
<p>i think it was 0.4 yeah…
not sure</p>
<p>guuuuys! the one of the difference of average of students thing… the one with the graph…</p>
<p>Was it average students or average OF the students??
is it 50 or 0?? what did you get??</p>
<p>someone tell me was it average students or average OF students…!!!</p>
<p>Someone who is 100% sure!!</p>
<p>2/5= 0.4 -_-</p>
<p>Now i remember yeah… that’s the answer i got…</p>
<p>if it was the average OF the students then is it
“It’s not 50?
Cuz the first place each year 100 new students are enrolled
And throughout 4 years , so the total will be 400
And the average 100
As for the second place, each year 50 new students are enrolled, throughout 4 years, =200
So the average is 50”</p>
<p>like someone previously said…
i’m confused…</p>
<p>that “of” makes all the difference… someone put me outa my misery…</p>
<p>btw, this is a previously administered test… march 2010…</p>
<p>here are the their scores:
<a href=“http://talk.collegeconfidential.com/sat-preparation/895510-march-2010-sat-results.html[/url]”>http://talk.collegeconfidential.com/sat-preparation/895510-march-2010-sat-results.html</a></p>
<p>@2200andbeyondXD I didn’t really get you! But the question asked to find the difference between the average number of students enrolled in the two schools. I am pretty sure that the answer is zero.</p>
<p>Holy s<strong><em>!! Why do this collegeboard give us previously administered tests?? F</em></strong> the bas****s!</p>
<p>and their math discussion is over here:
<a href=“http://talk.collegeconfidential.com/sat-preparation/880461-march-sat-math-thread-100.html[/url]”>http://talk.collegeconfidential.com/sat-preparation/880461-march-sat-math-thread-100.html</a></p>
<p>@hopingforbetter, iKNOW MMAN! THIS SUCKS… they are lazy idiots… nothing but that… pretty offensive i find it…</p>
<p>it’s the SAME EXACT BULLSH#$^!!!</p>
<p>I got it 0.2 
2/10 = 1/5</p>