$\noindent Hint: Expand using the Binomial Theorem to get $(2x^3 - \frac{1}{x})^n \equiv \sum_{k=0}^n \binom{n}k 2^{k}(-1)^{n-k}x^{4k -n}$. Consider $x^j$ for $j\in\mathbb{Z}$ where $j$ is an expression of $k,n$. When $j=0$, what does this say about $k$ and $n$? $
You could try the 2901 exam past papers just in case, as they seem to be similar difficulties (I think some questions overlap? or the old ones used to anyway.)
The first-year course packs are quite comprehensive (imo), so if you're willing to put in the effort to learn/understand it, then it should be manageable.
Draw the line y = 30-x (rearranging is x+y=30) on the cartesian plane. For x,y > 0, this is the first quadrant.
When you want the probability P(Y < 30-X), this is the double integral over the region bounded by the line x+y=30, x=0, y=0.
For double integrals, the outer limits are always...
Actually your part (ii) follows N(0,1) and in part (iii) it asks for the square of that. So your answer in part (iii) will also be a chi-squared random variable with parameter 1.
I'll do (v) which uses the result in (iv).
$\noindent (iv). If $X_i$ follows $N(\mu, \sigma ^2) $ in distribution, then $\frac{X_i - \mu}{\sigma}$ follows $N(0,1)$ (this is the same process as your other question above your latest post).
$\noindent (v). Define $Y:= \frac{X_i -...
If you guys use R I think it's likely that they'd use the functions
pnorm(a,mu,sigma) = some value, and this equals to P(X<a) for a r.v. with distribution N(mu,sigma^2)
So for that they would've given:
1 - pnorm(4.38,0,1) = some value
which of course equals P(Z>4.38) (since Z is usually...
Z is a random variable following the standard normal distribution and we denote it by writing
Z \sim N(0,1)
which means it has probability distribution function
f_{Z}(x) = \frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}} .
When you want something like P(Z>4.38), it's the area under the density...
Yes, in terms of passing, you'll have to at least get 50% in the final exams for both (iirc they aren't "double pass" subects and they were weighted around 50% of your final grade).
$\noindent Given that $\frac{\mathrm{d}r}{\mathrm{d}\theta} = \frac{3}{(1-r)^4}$ then taking the integral of both sides, and using the limits $r(0) = 0$, we have \\ \\ $\int_{0}^{r} (1-t)^4 dt = \int_0^\theta 3 ds$ (note that the variables of integration don't matter in this context, because...
I think you're allowed to use them without derivation. (Also, by derivation do you mean by the limit definition or by finding say dy/dx of sinx and taking the reciprocal?)