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Not sure what you mean by efficient, but yes, it is possible for a smaller volume object to weight more than a larger volume object – this happens iff the former object has a greater density than the latter. It is also possible for a smaller surface area object to weigh more than a larger...

It usually is post-trial (as in later than school trials).

The point of the exam is to give you a challenge.

I can't see the graph (actually, realised now it was for the function whose formula was given), but if the slope (namely [f(x) – f(3)]/(x – 3) ) approaches 2 from either side and f is continuous there (In fact, if [f(x) – f(3)]/(x – 3) approaches 2 (or some other finite number in general) from...

Re: HSC 2016 MX2 Combinatorics Marathon Actually, had a look on iTunes, and they're all free: https://itunes.apple.com/us/course/statistics-110-probability/id502492375 .

Re: HSC 2016 MX2 Combinatorics Marathon Yes, it is a good series. The full playlist is here: https://youtube.com/playlist?list=PL2SOU6wwxB0uwwH80KTQ6ht66KWxbzTIo . I think that lecture series is available on iTunes too.

Re: HSC 2016 MX2 Combinatorics Marathon I think this is called the de Montmort Problem or Matching Problem. It can be solved using inclusion-exclusion (which then can be used to prove the derangement formula in Paradoxica's link). It is worked through in this Stat110 Lecture from Harvard: ...

Re: MX2 2016 Integration Marathon Additional Q: $\noindent Find $\int \sqrt{\frac{6-x}{x-3}}\text{ d}x$.$

Re: MX2 2016 Integration Marathon I put the word corners in inverted commas because I was using it informally. I think for a cusp, the tangents should be equal for the two meeting branches (http://mathworld.wolfram.com/Cusp.html) (which is not the case for this function's graph).

Re: MX2 2016 Integration Marathon $\noindent Yes, this is essentially equivalent to Paradoxica's answer. This is because $\sqrt{1-\sin{\left(2x\right)}} = \sqrt{\left(\sin x - \cos x \right)^2}=|\sin x - \cos x|$, so $\frac{\sqrt{1-\sin{\left(2x\right)}}}{\cos{x}-\sin{x}}=\frac{|\sin x - \cos...

No, limit as h -> 0 of sin(1/h) does not exist. Best to use squeeze law. Limit as z -> 0 of 1/z isn't 0, but rather it does not exist. (1/z approaches +inf as x -> 0 from above and -inf as as x -> 0 from below.)

Re: HSC Chemistry Marathon 2016 You appear to be off by a factor of 10 compared to Flop21's reported answer. What method did you use? Did you try the method I posted (I have t actually typed numbers into a calculator but the method should be right)?

Well what they're asking you to show is essentially the fact that sin(x) has derivative cos(x). This is because a function f being differentiable at a satisfies f(a+h) = f(a) + h*f'(a) + o(h) (as h -> 0). Conversely, if a function satisfies f(a+h) = f(a) + h*A + o(h), where A is independent of...

Maybe the site doesn't have an AUD option.

It takes a bit of time for the view count to update I think.

For the second one, do you know whether any distributions were specified?

Haven't thought too much about the second part but my guess for it is: $\noindent Two numbers $X_1$ and $X_2$ (greater than 0) are randomly chosen. Then the fractional parts of $\ln X_1$ and $\ln X_2$ are computed and rounded to seven decimal places (call these $F_1$ and $F_2$ respectively)...

I assume for the first one: $\noindent There's a line of length $a$ units (where $a$ is a fixed positive real number). The points $P$ and $Q$ are chosen randomly (uniformly) on this line. Let $b$ be a fixed real number satisfying $0\leq b \leq a$ (I know he put $b < a$, but we can include...

Re: HSC 2016 4U Marathon $\noindent Remember though that there should be $i$ in the denominator, i.e. $\sin nA = \frac{z^{n} - \frac{1}{z^{n}}}{2i}$.$

Re: HSC 2016 4U Marathon Oftentimes once one has found a solution, they type it up without first searching (too hard) for other ones, because there's no guarantee that another one will be easily found and they want to get their solution up.