<p>What is the hardest problem you have ever encountered on the test?</p>
<p>This qn is pretty hard:</p>
<p>Which of the following has an element that is less than any other element in that set?</p>
<p>I. The set of positive rational numbers
II. The set of positive rational numbers r such that (r^2) is greater than or equal to 2
III. The set of positive rational numbers r such that (r^2) is greater than 4</p>
<p>a) None
b) I only
c) II only
d) III only
e) I and III</p>
<p>I don't understand the question :S</p>
<p>Yeah it looks pretty hard but I think the answer is (b) if my thinking is correct. Basically I think the question is asking "which one of these sets of numbers can have a number that is an absolute minimum for the set?" i.e., in a set of integers from [0, 3] 0 is the minimum. </p>
<p>For I, the smallest positive rational number is 0.
For II and III, any arbitrary rational number can be chosen for any arbitrary number greater than 2 or 4. There is no definitive minimum. For example, 1.42^2 = 2.0164 and 1.415^2 = 2.002225. This can go on forever as long as the number being squared is rational.</p>
<p>Now truthfully this took me some time to figure out and this problem would have definitely been a skip for me.</p>
<p>I think random questions about hyperbolas and their features are hard if they aren't fresh in your mind.</p>
<p>I guessed that the answer was B :D :D</p>
<p>If you don't know matrices, a question or two will trip you up. </p>
<p>Also, "least squares regression" will kill you if you don't know how to use your calculator properly.</p>
<p>Which one of these will change the standard deviation of some set of numbers?
I. Multiplying all the numbers in the set by a constant.
II. Adding a constant to each number in the set.
III. Subtracting a constant to each number in the set.</p>
<p>A) I Only
B) II Only
C) III Only
D) II and III
E) I, II, and III</p>
<p>yay AP Stats = best class ever</p>
<p>Which of the following has an element that is less than any other element in that set?</p>
<p>I. The set of positive rational numbers
II. The set of positive rational numbers r such that (r^2) is greater than or equal to 2
III. The set of positive rational numbers r such that (r^2) is greater than 4</p>
<p>a) None
b) I only
c) II only
d) III only
e) I and III</p>
<p>Sorry to correct you, but 0 is not a positive rational number, it is a non-negative rational number but just that.
The answer is (a) ... None</p>
<p>Reason - > For I . the lowest number if u consider 0.001, then 0.000001 will be smaller than that.. u can put infinite number of 0s and a 1 at the end. So no definite minimum number ...Or you can say that the range of the function is (0, infinity]
Note the open bracket at 0 which means 0 is not included.</p>
<p>for II, the smallest number is sqrt(2) but that is irrational, and u can have an infinitesimally small number than that, and u can go on and on but not reach a specific minimum..</p>
<p>For III , the minimum is clearly 2 but the equality is not given, it has to be greater than 2. So again III has no definite minimum.. (2.01, 2.0001, 2.000001) </p>
<p>Hence NONE are correct.</p>
<p>Which one of these will change the standard deviation of some set of numbers?
I. Multiplying all the numbers in the set by a constant.
II. Adding a constant to each number in the set.
III. Subtracting a constant to each number in the set.</p>
<p>A) I Only
B) II Only
C) III Only
D) II and III
E) I, II, and III</p>
<p>Since standard deviation is root mean square of the deviations from the arithmetic mean, additions or subtractions to the entities won't change the deviation, But multiplication will change the value of deviation by a factor of k (wtever constant u multiplied by).</p>
<p>The answer is (A).</p>
<p>For example:
3,4,5
avg is 4
deviations are 1, 0 , 1
Standard dev = sqrt(1^2 + 0^2 +1^2 / 3) = sqrt(2/3)</p>
<p>You will get the same, even if u take 4,5,6 or 2, 3, 4 ... simple.
But if u multiply any number, the deviations will change by a factor of the constant chosen.</p>
<p>It's funny how I forgot everything Math II related. :]
It's only been 3 months since I took it.</p>
<p>Well, the one I took was pretty simple; the only questions I didn't know were the topics I never covered in school, since I took the SAT after Algebra 2.</p>
<p>Therefore the hardest question I had was the one about matrices.</p>
<p>Something like:</p>
<p>a 2x2 matrice multipled by a 2x8 matrice gives what kind of matrice?
a) 2x8
b) 2x4</p>
<p>and so forth. But if you know it, then it's simple.</p>
<p>ya, when two matrix are multiplied (and in order to be able to be multiplied, they should have number of columns of first matrix = number of rows of second matrix) And the product matrix's configuration will be number of rows of first matrix X number of columns of second matrix.</p>
<p>So the product matrix will be of the type 2 x 8</p>
<p>wow these are all good problems. I especially like the rational number problem. That almost tripped me up cuz I neglected the positive. Lets keep the list going!</p>
<p>Do you guys think it's enough if i use Barron's to prep? I haven't had any statistics or trig courses.</p>
<p>Ok, I'll bite.</p>
<p>The units digit of 33^410 is:</p>
<p>A) 3
B) 7
C) 1
D) 9
E) 5</p>
<p>nice one but I wouldn't call it that hard. The patten is relatively simple 3^1 = 3, 2 = 9, 3 = 7, 4 = 1, 5 is back to 3. Thus 3^410 units digit is 9 or D. Barrons should be fine eritas44 but if you really want a challenge I suggest Meylani. He kicks the s<em>h</em>i*t outta yeh in a good sense:) Lets keep em coming.</p>
<p>Just ti-89 that problem. :P
I think the calculator helped me more than my brain did.</p>
<p>More questions please ;)</p>