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This thread was originally intended for that but ended up being littered with quite a few "joke questions" (if I recall correctly): http://community.boredofstudies.org/238/extracurricular-topics/349613/hard-questions.html .

Log based are generally restricted to be positive numbers (excluding 1), so yes.

Re: Statistics $\noindent If it's the union, we basically want the set of all points that belong to at least one of the $A_{k}$. The lower limit would be $0$, but the upper limit is $\infty$. So the union of those $A_{k}$'s is $\left(0,\infty)$.$ Edit: just noticed it was only k = 1 to 4. In...

Re: Discrete Maths Sem 2 2016 The ceiling function ƒ is not one-to-one, because there are two distinct values in the domain that get mapped to the same value by ƒ. For example, 1.2 ≠ 1.3, but ƒ(1.2) = ƒ(1.3) (= 2). (Remember, saying a function is not one-to-one is equivalent to saying there...

Re: Statistics $\noindent It is not the case that $\bigcap_{k=1}^{\infty}A_{k} = (0,4]$ if $A_{k} = \left(1 - \frac{1}{k}, k\right]$. We want the intersection (all points in common) of all the intervals $\left(1 - \frac{1}{k}, k\right]$ (for $k = 1,2,3,\ldots$).$ $\noindent The point $4$...

The set of all points equidistant to a pair of intersecting lines in the Cartesian plane is known to be the lines that pass through the point of intersection and bisect the angles between the lines. In this case, this gives the lines y = ±x.

Re: Statistics To find Pr(X > 1/3) for that one, we'd just integrate the PDF of X (f(x) = 2x) from x = 1/3 to x = 1.

Re: Statistics You can do it using that conditional probability formula you wrote, you'll just need to calculate each of the terms that appear using the given PDF.

Re: Statistics If there are two values of X that yield the same value for Y, then the probability of Y taking that value is just the sum of X taking on each of the values that gave that Y value. The probability entry for Y = 2 should be 1/4, because Y = 2 occurs iff X = -1 or X = 1, and Pr(X...

Re: MATH1131 help thread When x is large, f(x) "looks like" x/(2x^3) = 1/(2x^2), because the -1 on the denominator becomes negligible. So we know we want to compare f(x) to 1/(2x^2) (or can just do it to 1/x^2, since all we need is the limit of f(x)/g(x) to be a positive constant), and this is...

Re: Statistics You would need to find the range of Y based on the range of X. Also, the probability Y takes on some value would be easily to see from the probability distribution of X. For example, if X took the value 1 with probability 1/2 and 0 with probability 1/2, then Y would take the...

Re: Statistics That's the conditional probability that X equals 0 given X < 1.

Re: MATH1131 help thread $\noindent The limit comparison test helps us decide whether a given improper integral is convergent or not by comparing the integrand to another function whose improper integral we know is convergent or divergent. For example, in question (a), let $f(x) =...

Re: Statistics $\noindent If $X$ and $Y$ are discrete, then if you know their joint pmf $f(x,y) = \mathbb{P}\left(X = x,Y = y\right)$, we would have $\mathbb{E}\left[XY\right]$ be given as a double sum over all values in the range of $X$ and $Y$, that is,$ $$\mathbb{E}\left[XY\right] =...

Once you've done a), b) is easy (just put h = 0.01 in your formula from a)). To get the answer to c), take the limit as h -> 0 in your answer for part a).

Now that those are cos(theta) etc., all you need to do is find the intersection of two lines. This can be done by solving for x in one of the lines (in terms of y), and plugging this into the other line's equation (i.e. standard procedure of solving two simultaneous linear equations in two...

Find all values of x such that f(x) > 6.

What would be the x-intercepts, given your factorisation?

You don't need to do it that way. Note that the condition pq > 0 gives us precisely two possibilities: either p > 0 and q > 0; or p < 0 and q < 0. If p > 0 and q > 0, we have px^2 + q > 0 for all real x (since x^2 is non-negative for real x), so the quadratic has no real roots. If p < 0 and q...

Consider cases of a > 0 and a < 0.