<p>To Noober: I'd keep studying. I did two SAT Math practice tests, got 800's on both, and stopped studying. It ends up that every time, I got the questions like the first one posted wrong... so, one the REAL SAT, I got all of those wrong (plus, my calculator broke, and I realized upon reading about the answers in the SAT threads that I added some things incorrectly. Gaaah). The result? A 620. That was definitely a blow to my confidence.</p>
<p>I'm going to prep much more for the Math II... maybe a good score will convince colleges that I don't suck at Math =P</p>
<p>Does anyone know anything about Kaplan's Math IIC because on the practice tests, i've only been getting 620s and i'm really scared i'll get even lower in the real SATII</p>
<p>i took mathiic in october, having not yet taken precalculus, and scored 760. I'm retaking it on saturday, having now finished precalc, and I can't decide whether it's worth bothering to study.</p>
<p>I don't remember what was on the test at all. Does anyone know if precalc will have covered all the loose ends or need I go find a prep book? :S</p>
<p>Combinations, permutations, and the normal types.</p>
<p>Normal types would be:</p>
<p>1.There are 3 shirts, 2 pants, and 4 hats. How many different outifts are possible if an outfit has pant, shirt, and a hat? You would do 3<em>2</em>4 to get 24 different outfits.</p>
<ol>
<li><p>You have 3 coins. What is the probability of getting two heads and one tail if you flip all three coins once? You do this by first finding the number of outcomes (2 x 2 x 2) which is 8. You know that 2 coins have to be head and one must be tails. Seeing all the possibilities involving 2 heads by making one coin tails. This happens 3 different ways. Therefore the answer is 3/8.</p></li>
<li><p>You have 2 concentric circles. One has a radius of 2. The other has a radius of 4. If you pick a point at random, then what is the chance of choosing a point in the shaded region (shaded region is outside the first circle but inside the second circle)? You do this by finding the areas of both circles which is 4pi and 16pi. 4pi/16pi=1/4 chance of hitting in the small circle. The answer is therefore 3/4.</p></li>
</ol>
<p>Can anyone show me how to answer this question. Barron's doesn't explain it that well.</p>
<p>A craftsperson has six different kinds of seashells. How many different bracelets can be constructed if only four shells are to be used in any one bracelet?</p>
<p>You do 6 Cr 4. You can do this with your calculator by hitting 6 then math>PRB>nCr then 4. This is a combination type problem where order does not matter.</p>
<p>In permutation problems, order DOES matter. For example, 3 cars are in line at a drive thru. How many different ways can the first 2 cars be arranged? You would do 3 Pr 2.</p>
<p>To do combination and permutation by hand:</p>
<p>a Pr b= a!/(a-b)!</p>
<p>a Cr b=a!(a-b)!(b)!</p>
<p>! is the factorial sign. FOr example 4! is 4x3x2x1=24</p>
<p>This problem requires you to find the number of FREE circular permutations because flipping the bracelet is allowed.</p>
<p>So first find number of different compositions of bracelets (i.e. without caring about order). 6 Cr 4 is 15. Now use the formula for free circular permutations: Pn=(n-1)!/2 In this case, n=4. So do 15(3!)/2=90/2=45</p>
<p>The Barron's explanation accounts for the dividing by two. Imagine a bracelet that is red/blue/green/yellow, and imagine one that is yellow/green/blue/red. They are the same if you flip the second one over. The computation accounts for both of those, so you have to get rid of one for each by dividing by two.</p>
<p>I don't know what the 3! is about, or the whole computation thing.</p>
<p>Ok, I got the answer, but I still am a little confused about the permutation.
6 nPr 4 = 360
Divide this by four because a bracelet can be rotated into 4 different views, but those views are still of the same bracelet.
Divide by two because the bracelet can be flipped over and still be the same bracelet.
360/(4*2) = 45</p>
<p>However, the permutation part doesn't make sense, because it seems to me that these bracelets would be allowed to have repeated shells. (like red, blue, red, green, or even having all one color), but the permutation doesn't allow for this. If the question is meant to exclude repeated shells, then the permutation above works.</p>
<p>I dunno, I didn't do it the way the books says, so maybe I just found the same answer coincidentally.</p>
<p>Crap, ok, my problem was that that problem is horribly written.
1) I was thinking bracelet without a clasp: with a clasp it is basically just linear that can be flipped, without a clasp it is circular and can be rotated to make the same bracelet.
2) It didn't specify that no shell could be repeated. It said 6 different KINDS of shells, implying that there is an abundance of each, and they can be repeated.</p>
<p>They had better not put such an ambiguous question on the actual test.</p>
<p>I have a question, when do we need to know how to use synthetic division?? I've always been using my calculator to find zeros so I don't know if I should review this...</p>
<p>Also, how do we find imaginary factors?????</p>
<p>Oh, and do we need to know Descartes' Rule of Signs?</p>
<p>Just know that imaginary/complex factors come in conjugate pairs; if 2+4i is a factor, so if 2-4i
Also, they probably won't ask about Descartes' Rule of Signs, but you could always use your TI-8_'s "zero" function to find the zeroes.</p>