<p>@kristiuna, the “best fit” line is defined as the line that minimizes the sum of the squares of the errors (you’ll probably learn this in a statistics course). One way to think of it, for every 5 ounce increase, the price increases by 60 cents or so. So the average cost per ounce is about 12 cents.</p>
<p>guys, I have this question here in the Official SAT Test 3 Sec. 5 no. 6. It’s been asked a lot, but there’s just one thing I don’t get about it and I didn’t get answer from all the threads it was asked on previously:</p>
<p>If x does not equal 0 and x is inversely proportional to y, which of the following is directly proportional to 1/(x^2)?</p>
<p>A) -1/(y^2)
B) 1/(y^2)
C) 1/y
D) y
E) y^2</p>
<p>The answer is E. I know the steps to it:</p>
<p>first, since x = 1/y then y = 1/x so if we square the whole thing we get y^2 = 1/x^2 and they say this is the answer.</p>
<p>My question is: isn’t y^2 = 1/x^2 an INVERSE statement? How on earth could this be direct proportion. It should be 1/y^2 = 1/x^2 for it to be directly proportional. Right?</p>
<p>@satphro227 thanks a lot for the clarification!! I get it algebraically now but I now need to understand it visually or on a more abstract basis. Dr. Steve, are you there? :D</p>
<p>@jasonjackson789,</p>
<p>y^2 is inversely proportional to x^2.
y^2 is directly proportional to 1/(x^2).</p>
<p>Hope that clears things up.</p>
<p>@rspence thanks for the reply, but I still don’t get it:</p>
<p>y^2 is inversely proportional to x^2 means y^2 = 1/x^2 ----> how could this statement be both directly and inversely proportional at the same time?</p>
<p>y^2=c/x^2 means y is inversely proportional with x^2. Let’s say I called 1/x^2=a. then y^2=ca. Here y^2 is directly proportional with a so y^2 is proportional with 1/x^2.</p>
<p>ok, I get that part. However, let’s try to plug real numbers, and let’s keep k = 1:</p>
<p>let y = 2, then 2^2 = 1/x^2 which gives x^2 = 1/4 and y^2 = 4 —> inverse relation</p>
<p>let y = 3, then 3^2 = 1/x^2 which gives x^2 = 1/9 and y^2 = 9 —> still inverse relation</p>
<p>See why I don’t get it? Because when I plug numbers, it gets out as an inverse relation…</p>
<p>With k = 1 we have xy = 1.</p>
<p>If y = 2, then x = 1/2, and so 1/x^2 = 4 and y^2 = 4.
If y = 3, then x = 1/3, and so 1/x^2 = 9 and y^2 = 9.</p>
<p>So it looks like 1/x^2 and y^2 are directly proportional. The constant of proportionality is 1.</p>
<p>I still don’t get it 
I get 750-800s on all my practice tests but this one is utterly stupid. Okay, when I try to reason it out, here’s how it goes: x^2 = 1/y^2 (suppose that k = 1)</p>
<p>then isn’t the inverse of an inverse, a direct relation? So the above one is an inverse relation. Then what’s the inverse of that? Well, let’s just change one of the sides, so we make 1/x^2 = 1/y^2 —> direct proportion. I mean in the first one (the supposedly correct answer) if we cross-multiply then it becomes x^2 * y^2 = 1 —> an inverse statement. But if we cross-multiply in the second one, it becomes x^2/y^2 = 1 —> a direct statement. How could this be wrong!! It goes against all what I’ve learned about variations :S</p>
<p>I mean your first statement in your answer “With k = 1 we have xy = 1.” is an inverse statement!!</p>
<p>I don’t see how you’re thinking it…y^2 is proportional to the quantity 1/x^2. So if 1/x^2 is multiplied by some constant, y^2 will also be multiplied by that constant.</p>
<p>2jason – you almost have it, and it’s staring you right in the face!</p>
<p>You said:</p>
<p>let y = 2, then 2^2 = 1/x^2 which gives x^2 = 1/4 and y^2 = 4 —> inverse relation</p>
<p>let y = 3, then 3^2 = 1/x^2 which gives x^2 = 1/9 and y^2 = 9 —> still inverse relation</p>
<p>And you were right! You showed that y^2 varies INVERSELY with x^2.</p>
<p>But you have one more step to take: the problem asked what varies DIRECTLY with not x2 but ONE OVER x^2.</p>
<p>Using your numeric examples, but remembering to invert x^2, you’ll see that y^2 comes out the same as 1/x^2.</p>
<p>NOTE: rspence and Dr Steve both have it right and simpler. I just wanted you to see how close you were to having it right your way.</p>
<p>@pckeller: THANK YOU THANK YOU THANK YOU!!! I get it now completely!! I now see why I was thinking about it in the wrong way!! Man, now that’s one subtle difference! Thank you too Dr. Steve and everyone else too for helping me out with this one :D</p>
<ol>
<li>If x and y are numbers such that (x+9)(y−9)=0,
what is the smallest possible value of x2+ y2?
(A) 0
(B) 9
(C) 18
(D) 81
(E) 162
S T</li>
</ol>
<p>The median age of a group of 15 students is 17 years. If the oldest student in the group is 20 years old, which of the following could be the number of 17 year olds in the group?</p>
<p>I 3
II 10
III 14</p>
<p>A I only
B I and II only
C I and III only
D II and III only
E I, II, and III</p>
<p>The correct answer is E, but how do you get to this answer? Is the oldest student being 20 years old have any purpose in answering the question?</p>
<p>The oldest student being 20 doesn’t matter too much here. Just make the 8th number 17. Now you have a lot of flexibility with the other numbers. You can make them all 17 (except the last one which must be 20). You can make 3 of them 17, 6 of them 1 and 6 of them 20. You can make 10 of them 17, 3 of them 20, and 2 of them 15.</p>
<p>If (x + y) (x^2 - y^2) = 0 , which of the following must be true?
(A) x = y
(B) x = -y
(C) x^2 = y^2
(D) x^2 = -y^2
(E) x^3 = x^3</p>
<p>C is the correct answer, but what makes B more incorrect than C? This was the final question on section 2 of the 2006 Saturday PSAT.</p>
<p>B is incorrect because x could equal y, therefore B does not have to be true. Here, x^2 = y^2, or x = +/- y, answer is C.</p>
<p>a train started at station s and has been traveling r miles per hour for t hours in terms of r and t ,how many miles from station had the train travled 3t/4 hours ago ? the answers is (E) i cannot understand why it is e</p>
<p>^ would be helpful to list answer choices, (esp. E). </p>
<p>I assume the answer goes something like this: at 3t/4 hours ago, the train has traveled for t-3t/4 or t/4 hours, so distance = rate * time, so distance = r * t/4 = rt/4 miles from station.</p>
<p>thanks for your help</p>