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That's pretty much 0 (if Z is standard normal).

I don't know if this was the reason why, but in the Q you typed above, there's a typo (top-right entry should have 6 rather than 5 in the square root).

Yes

$\noindent If $A = PDP^{-1}$ in general ($D$ diagonal), then defining $B= PD^{\frac{1}{2}}P^{-1}$, we have $B^{2} = A$. Here $D^{\frac{1}{2}}$ is a square root of $D$, which is a diagonal matrix with diagonal entries all being square roots of the entries in $D$. In your example, you can do this...

Yeah but the third derivative was negative, so it'd be the negative of what you wrote there.

It won't be -x^3. It'll be a different cubic. It will be an odd cubic that satisfies f(pi) = 0.

You could probably also do it by computing the second partial derivatives at (x,y) ≠ (0, 0) and investigating their limit as (x, y) -> (0, 0), but there's no need to do that since you've already computed the mixed second-partials at the origin in an earlier part of the question (part b)).

It didn't work because the hypotheses of Clairaut's Theorem were not satisfied by that ƒ (I assume you know what typical hypotheses are for it).

It looks like your log term was not differentiated correctly.

$\noindent Those are the same. Since $\overline{X} = \frac{\sum\limits_{i=1}^{n}X_{i}}{n}$ (definition of the sample mean), we have $\frac{1}{\overline{X}} = \frac{n}{\sum\limits_{i=1}^{n}X_{i}}$.$

We can just prove it by cases: either X ≥ Y occurs or Y > X occurs, and in either case the LHS is ≥ 0, since f and g are increasing functions. Same is true (replacing X by x and Y by y) if x and y are just real numbers in the domain of f and g if f and g are functions defined on a subset of the...

$\noindent We have $\sin(x+y) = \underbrace{\sin x}_{= \frac{2}{3}} \underbrace{\cos y}_{= \frac{3}{4}} + \cos x \sin y$.$ $\noindent We know the values of $\sin x$ and $\cos y$. From the given information alone, we can only pin down the value of $\cos x$ and $\sin y$ up to a $\pm$ sign...

$\noindent Well for one thing, if $T(\mathbf{v}) = \mathrm{proj}_{W}(\mathbf{v})$ for all $\mathbf{v} \in V$, we need that $T(\mathbf{v})\in W$ for all $\mathbf{v} \in V$ (because by definition $\mathrm{proj}_{W}(\mathbf{v})\in W$). Now as we know, a subspace of $V$ that contains $T(\mathbf{v})$...

Claim: W := im(T) is such a subspace. Proof: Exercise.

Re: UNSW MATH1081 Discrete Maths If we know the no. of edges is n(n-1)/4, since the no. of edges is an integer, what does this tell us about n?

Re: Statistics $\noindent It would be because the distribution of the test statistic is (or approximately is) normal, so a $95\%$ confidence interval would be the estimate plus-or-minus $1.96$ times its standard error. This comes from the fact that a standard normal distribution has 97.5-th...

Re: Statistics It's a standard formula for the variance of a sample proportion, and basically comes from the formula for the variance of a Bernoulli(p) random variable (or variance of a Binomial(n, p) random variable). It's worked through on this page, for example...

Re: Statistics Remember, there is an identity Var(cX) = c2 Var(X), if c is a constant. In other words, to get the variance of a constant times something, the constant can come outside but gets squared.

Yes, it is possible.

Re: Statistics Yes, this is the case.