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Re: HSC 2015 3U Marathon Did you leave the 's' out of 'fat'? (Or perhaps it is 'r' ?)

Re: HSC 2015 3U Marathon So you haven't done Ext 1 Projectile Motion yet?

Re: 2015 permutation X2 marathon Fibonacci (n+2)

You didn't say "the Turkish government". You said Turks. Perhaps a Turk could look at the racist policies of Australian governments past and present, and use that to claim "Australians are shitbags". Would that be a fair statement? And I didn't say you were totally wrong about the housewives...

Whoever wrote that should go on detension.

How about 3D trig involving Parameters and Related Rates.

Re: 2015 permutation X2 marathon p and q are real numbers. p is randomly chosen on the interval -a ≤ p ≤ a q is randomly chosen on the interval 0 ≤ q ≤ b What is the probability that the equation x² + px + q = 0 has real roots ? [Note: This is technically a 2 unit question]

His what is implying it?

Except that was because they put probability last in the Ext 1 exam. I could be wrong, but I don't think they'd go straight from last in Ext 1 to last in Ext 2.

Yes it IS just doubled. But you are not doubling whole marks. Your marks are recorded to at least one decimal place (it might be more). The unrounded mark gets doubled. No, the ATAR is calculated on pre-aligned marks.

Re: 2015 permutation X2 marathon That was my worry too. I know the answer is correct, but I'm not sure how to justify that, other than it seems to be common sense (and yes, I know a lot of things that seem to be common sense turn out not to be true).

If you count all mass shootings in the US (more than one dead or injured) there have been 297 so far this year. If you count shootings with more than one death, there have been 84. The total number of gun deaths in the US this year is 9995. That's one every 40 minutes. 9996 .... 9997 ..... 9998...

Re: 2015 permutation X2 marathon Let Pn be the probability that the man goes broke, given that he currently has $n. To go broke starting from $1, he must either: (1) lose $1 with probability (1-p); OR (2) gain $1 with probability p AND then go broke from a new starting point of $2...

Assuming you are saying that the problem with stereotyping is making a sweeping generalisation (because I can't think of any other possible problem with it - in fact, stereotyping is just a commonly used generalisation about people): Didn't you say: "Highest anti-vax rates are in rich suburbs...

Just like all Ext 2 texts, it has its good points and its bad points. It's too late to think about another book now (you should be doing past papers) but you really need two or three books to cover the course properly.

Re: 2015 permutation X2 marathon Although this is related to series, you don't need to sum an unspecified or infinite number of terms. No solution yet, but I'll give the answer: For p ≤ 1/2, the answer is 1 (as Integrand already suggested) For p > 1/2, the answer is \frac{1-p}{p}...

Re: 2015 permutation X2 marathon Not correct ... but at least it is a sensible answer - it gives you what you would expect for p=0 and p=1. I can't comment further without working or some idea of what you were thinking. But ... it's bed time for me.

Re: 2015 permutation X2 marathon A man has $1 to his name. Every day he either gains $1 (with probability p) or loses $1 (with probability 1-p). This stops if and when he goes broke. Find, as a function of p, the probability P that the man will eventually go broke.

Re: 2015 permutation X2 marathon Thanks for that. I hope you don't mind if I stick with my version - it seems a little more intuitive - I can practically "see" it. In fact, I am wondering if there is a geometric connection to Archimedes' result.

Re: HSC 2015 4U Marathon Ah ... no. I just found it in the '74 paper. At that stage I was probably practising my times tables.