Search results

What were your thoughts on this year's exam?

Entirely depends on what "part thereof" meant. It seems to me it should just mean you get charged 54 cents for each 30 second chunk, as long as you enter a 30 second chunk (but again, I don't do General Maths. But this is what I thought 'part thereof' meant from general knowledge.). So something...

Yeah thought so: in other words (assuming this is the case), the thing I missed is that the "mega" in megabits is 1000^2, whereas in megabytes it's 1024^2 (I was assuming "mega" referred to the same factor in both cases). Looking briefly at Wikipedia, there seems to be some ambiguity about...

(Assuming you're talking about Question 30 (b) (ii)) How did people get 61? (Caveat, I don't do General Maths though, so I might have missed something.) Note that 1 byte = 8 bit. So 3072 MB = 3072 megabyte = 3072*8 mb (megabit). Therefore, the speed in megabit per second required (noting 7 min...

This method would unfortunately fail to exclude volume from the parts of the rectangle's "edges" where there is no chocolate. These should be excluded (since there's no chocolate there).

$\noindent The area of each shaded region in the `top view' is $5^{2} - \pi \times (2.5)^{2}$ (proved below). There are six such shaded regions and the depth of each is $6$ cm (since there are two levels of $3$ cm high marshmallows). So the volume of chocolate used is: $6\times \left(5^{2} - \pi...

It is 193.1417... (cubic cm), which is roughly 193 to nearest whole number.

Yeah it's $0.

If I recall correctly, we can't link the ATAR Notes site here. But anyone curious to see the paper could probably google that site and find it (if the paper's been uploaded there – haven't checked myself).

What were the equations? Note that if we're given simultaneous equations and come up with some answer, we can (and should) test it by subbing it into the equations and seeing whether they all hold. If we have a calculator with us, this is pretty easy to do.

Re: HSC 2017 General Mathematics Marathon Not sure if you can't see it too, but basically the triangle they gave is impossible. They should have just excluded the part saying that C is due East of A, because without that, the triangle is already uniquely determined by the sides 18 and 12 and...

Re: HSC 2017 General Mathematics Marathon I can't see the above LaTeX at the moment. Test: $\noindent test$

Re: HSC 2017 General Mathematics Marathon $\noindent That's an impossible triangle. We can't have all those angles and lengths and also have $C$ be due East of $A$. If we did have all these conditions, then the height of the triangle is $18\sin \left(90^\circ - 32^\circ\right)$ and also...

This is pretty much all rubbish, of course.

I think Carrotsticks said the 2016 one was designed to be generally easier than the earlier BOS Trials.

Re: HSC 2016 MX2 Combinatorics Marathon $\noindent A group of $n$ boys numbered $1,2,\ldots, n$ ($n\geq 3$) sit at a round table in that order. Person $1$ starts with a counter and passes the counter at random (with $50-50$ chance) to one of the two people sitting directly next to him. The...

Re: ATAR Notes vs Bored of Studies Possibly because it's a competing site. Or there might be a deeper reason.

Since we needed to prove something equalling (2^{2n} times) cos(n*pi/2), it's natural to make arccos(x) equal pi/4, so that the 2n*arccos(x) becomes n*pi/2.

The only subject you can't get a good ATAR without is English.

Re: MATH1231/1241/1251 SOS Thread The question's asking for a distribution, the answer isn't a number. Hint: hypergeometric distribution (recall this models sampling without replacement. Check Wiki for more details.).