Search results

Re: Discrete Maths Sem 2 2016 Here's some hints/sketch. Note (why?) that we really just need to prove the claim for proper fractions, i.e. numbers of the form p/q, where p, q are positive integers and p < q (so fractions less than 1). The general case will follow easily from this. So, let p...

Re: Discrete Maths Sem 2 2016 That basic idea is hinting at using induction on n.

$\noindent 1) A point $P(x,y)$ lies on the locus iff $|y| = 2|x|$ (since $|y|$ is the distance of $P$ from the $x$-axis, and similarly $|x|$ for the $y$-axis). So the locus is the pair of interesting lines $ y = \pm 2x$.$ $\noindent 2) A point $P(x,y)$ lies on the locus iff $|y| = |x|-3$...

Isn't it a bit unrealistic to expect to self-learn a whole HSC course (plus do past papers on top of that) in the holidays? You'd probably burn yourself out.

$\noindent You may find it easier to obtain the $P_n = (2n-1)\times (2n-3)\times \cdots \times 3\times 1$ formula by working forwards rather than backwards. Note that $P_1 = 1$, and then using the formula from part a), $P_2 = 3P_1 = 3\times 1$, then $P_3 = 5P_2 = 5\times 3 \times 1$, and we see...

$\noindent Note that clearly $P_1 = 1$ (only one way to form the two-player team if there's only one pair to begin with). Now, we know $P_n = (2n-1)P_{n-1}$ for $n\geq 2$. Just keep plugging back into the recurrence, so $P_n = (2n-1)(2n-3)P_{n-2}$, since $P_{n-1} = (2n-3) P_{n-2}$. So $P_n =...

Auxiliary angle method is something required to be known by 3U students, isn't it?

$\noindent Note that $P_n$ is the no. of ways to assign two-player teams when there are a total of $2n$ people, for $n \geq 1$. So to form the recurrence, note that for a particular person in the group of $2n$ people (say Alice), there's $2n-1$ ways to assign their partner, and once this is...

Yes. The third moment has to do with skewness (which is more precisely the third standardised moment). The fourth moment has to do with kurtosis (which is more precisely the fourth standardised moment). Skewness intuitively has to do with how asymmetric your distribution/data is about its mean...

Do you know why they still get listed as getting the University Medal?

$\noindent We have $\mathbb{P}\left(X_1 > y \text{ and } X_2 > y \text{ and } \ldots \text{ and } X_5 > y \right) = \mathbb{P} \left(\bigcap _{i = 1}^{5} X_i > y\right) = \prod _{i=1}^{5} \mathbb{P}\left(X_i > y\right)$, since the $X_i$'s are \textbf{independent}.$ $\noindent (Remember the...

$\noindent The work done is given by $W = mgh$, where $m$ is the mass of the brick (4.0 kg here), $h$ is the vertical distance it is lifted (1.5 m) and $g$ is the acceleration due to gravity $\big{(}$9.8 $\text{m s}^{-2}$$\big{)}$. Subbing these numbers in, $4\times 9.8 \times 1.5 = 6 \times 9.8...

If you want to actually see a proof, you can find one here: http://math.stackexchange.com/questions/221078/poisson-distribution-of-sum-of-two-random-independent-variables-x-y .

Yeah that's a true statement (assuming those are independent r.v.'s). That says that a sum of independent Poisson random variables is Poisson with rate parameter equal to the sum of the original rate parameters. It was proved by leehuan as an old homework exercise I believe (but the proof isn't...

Yes, it is possible for you to get 85 ATAR.

$\noindent Well yeah, suppose person A gets internal of $a$, with B and C getting $b$ and $c$ respectively for internal, and $a > b > c$. If in the externals these are reversed, then person A gets a HSC mark of $\frac{a+c}{2}$, B gets $b$, and C gets $\frac{c+a}{2}$. So C would always get the...

Depends on the marks (they clearly don't just average those ranks).

Yeah (it's even theoretically possible to come first in neither externals nor internals, and still top the state).

It's very much possible to top the state in a subject without being first in your cohort internally (especially if you're from a top-ranked school).

Which part were you unsure of?