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Sorry, I should have put a bit more. It's a pretty similar thing to what 香港! did. Going from where you got up to: I_n = n/2∫x²(1-x)^n-1dx I_n = n/2∫x²(1-x)^n-1dx + n/2∫x(1-x)^n-1dx -...

Both add and subtract n/2∫x(1-x)^n-1 dx. Take out a common factor of (1-x) after combining the negative one with what you already had and you will get I_n-1 and I_n.

Let the top of the mast be H and the point vertically below it be D. In ΔABH we have sin30°/BH = sin15°/90 ---> BH = 90sin30°/sin15° In Δ BDH (right angled) we have sin45° = h/BH ---> BH = h/sin45° where h is the height of the mast. combining both...

Hmm, if person 1 was born on a leap day then the probability that they share a birthday is 1/1461. If person 1 wasn't born on such a day then the probability is 4/1461. To factor both in then you need to find P(both born on a leap day)and P(both are not born on a leap day). Giving you...

Most questions seem to ignore leap years but add it in if you want.

Correct me if I'm wrong but wouldn't the easiest way be: Person 1 has a birthday The likelihood that person 2 has this certain birthday = 1/365 ∴ P(same B'day) = 1/365

Hmm, that's pretty cool. I've seen the e^x power series before (I have a pretty rough understanding of how it comes from the taylor series) but I didn't notice it here. Thanks for pointing it out.

It would probably be sin(πx) because (sinπ )x = 0 which would kill the question.

http://www.geocities.com/fourunitmaths/sgs2004.pdf 8 bi) At the start of this you're establishing an inequality under the condition that 0 < x < 1 .... I'm not too hot at inequalities but from this I believe it follows that: a) 0 < xⁿ < 1 b) 0 < (1 - x)ⁿ < 1 c)...

So far I've put in applications for: UNSW Newcastle Melbourne Monash UTas UWA UQ

Anyhow, that one bugged me for a bit. I'd like to see someone come up with a more elegant solution for it. Btw, I think when I calculated the values of a couple of the U₄/U₅ terms I may have been unco and stuffed them up (after saying that part was easy :p).

I've come to that answer a couple times using shaky-common sense but not much of it had strong mathematical grounding. This is the best I could come up with but it's still a little shonky: Consider the case with (k+1) letters and (k+1) envelopes... there are 'k' letters which can go into...

vi) Deduce that &sum; (k = 2 to n) ⁿC_k U_k = n! - 1 What this is doing is summing up all the individual cases: When k=2, ⁿC₂ U₂ = the number of ways in which the letters can be arranged where two letters are out of place...

There may well be better ways to do this but this seems to work. If Z = cos@ + isin@ then 1/(1+z) is equal to, 1/[ (1 + cos@) + isin@) * [(1+cos@)-isin@]/[(1+cos@)-isin@] (to make the denominator real) = [cos@ +1 - isin@]/[2cos@ + 2] = 1/2{1 - i(sin@)/(cos@ + 1)} now...

a) The total number of ways to arrange 5 distinct letters into 5 distinct envelopes is 5! = 120 # ways to put all letters in incorrect envelopes = U₅ P(all incorrect) = (ways to arrange them incorrect)/(total # ways to arrange them) &there4; P(all incorrect) =...

i) This is pretty simple. If you have two letters and two envelopes there is only one way to get them messed up. If you have three letters and three envelopes lined up (for the sake of visualization) then 2 letters can go in the first envelope, 1 in the next and 1 in the next. ii) Let me...

If you let Z₁ = cis&theta; and Z₂ = cis&phi; then Z₁.Z₂ = cos&theta;cos&phi; - sin&theta;sin&phi; + i(sin&theta;cos&phi; + cos&theta;sin&phi; ) = cos(&theta; + &phi; ) + isin(&theta; + &phi; ) = cis(&theta; + &phi; ) &there4...

y = (x-4)/(x-1) so switch the x's and the y's ... x = (y-4)/(y-1) yx - x = y - 4 y(x - 1) = x - 4 &there4; y = (x - 4)/(x - 1) so f(x) = f^-1(x) As acmilan said, it is it's own inverse.

I suspect they mean sydney high.

On our second last day we had a private year 12 assembly with Gough Whitlam where he spoke to us and where there was time for a few questions. Later that day we had a graduation assembly with the school (which was also a farewell for out grade deputy and our deputy year adviser). A lot of...