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It's essentially a maths/finance degree but with an honours year attached to the end. You don't really learn anything special and certain subjects in the program (SCIF and MATH2881) aren't great. You have greater flexibility in a maths/finance degree while doing similar courses. The name is just...

As said before you are better off doing a maths major if you like maths. Alternatively, if you enjoy physics as well and don't mind some hard work, do engineering. To be honest, you don't really do much maths in any commerce major.

Quant Risk (UNSW) is pretty overrated. Don't do it unless you are keen on honours.

That's the bare minimum. In finance they usually take the very top of the finance students applying (i think top 20 in cohort). So you would be hard pressed to find a supervisor (a good one anyway) if your WAM is less than 85. Finding a supervisor willing to take you on is also part of...

Can't part 2 be done by simply applying differentiation through first principles. Eg. let x=a+h, y=a for h>0, then for all a: f(a+h)-f(a)&\leq h^2\\ \frac{f(a+h)-f(a)}{h}&\leq h\\ \lim_{h\rightarrow 0} \frac{f(a+h)-f(a)}{h}&\leq \lim_{h\rightarrow 0} h\\ f'(a)&\leq 0 Then choosing for x=a...

Its similar to how you split the cases in the first place: 1) Find all the cases where 1 suit is missing (this includes where 2 and 3 suits are missing): - Choose a suit: 4 - Choose 8 cards from remaining 39 In the above case, you will notice the cases with 2 and 3 suits missing will be double...

This

^Nope, its possible in any 4+ year degree combo with commerce

The flaw in your first attempt comes from steps 4/5- where you choose other cards from the remainder. It means you will double count some cases. For example: Suppose in case where hearts and diamonds are missing: 1) choose 10C and 10S 2) Choose 6 other cards to be 9S,9C,1S,2S,3S,4S This is...

This is a very nice probability question Replace heads with tails and Alice winning to Alice losing, then the answer becomes obvious by symmetry. (i.e. for Alice to lose, she needs to flip strictly more tails than Bob does and P(Alice lose)=P(Alice wins) by symmetry).

hschard is on the right track. P(hitting bulleyes)=1/5=2/10 P(hitting target)=9/10 So P(hitting target but not hitting bulleyes)=9/10-2/10=7/10 So: P(hitting 2 bulleyes and 2 targets(but not bulleyes))=(7/10)^2*(2/10)^2*4C2 P(hitting 3 bulleyes and 1 targets(but not...

Alternatively, if you really want to do actuarial studies, do it now by transferring to com/science degree and then get a one year diploma in education afterwards for qualification to teach maths.

(1) Boy girls alternate It will either be BGBGBGBG or GBGBGBGB Arranging the boys in either case will be 4!. Same with the girls. So number of arrangements = 4!*4! (number of arrangements for each case) * 2 (number of cases) (2) 2 girls together: Treat 2 girls ("GG") as one group. Then...

P(Not winning jackpot in 1 draw)=59/60 P(Not winning any jackpots in n draws)=(59/60)^n=1-P(Winning at least 1 jackpot in n draws) So solve 1-(59/60)^n=0.98 for n

\int_0^{\frac{\pi}{2}} (cosx)^{2n}cosxdx &= \int_0^{\frac{\pi}{2}} (1-sin^2x)^n cosxdx\\&=\int_0^1 (1-u^2)^n du \\&=\int_0^1 \sum_{k=0}^n ^nC_k (-u^2)^k du \\&=\sum_{k=0}^n ^nC_k \left[\frac{(-1)^ku^{2k+1}}{2k+1}\right]_0^1\\&=\sum_{k=0}^n ^nC_k \frac{(-1)^k}{2k+1} edit: see above ^^

You mean finding \lim_{x\rightarrow 0} \frac{e^x-1}{x}=1 without L'hopital's? Then try using the Taylor expansion of e^x=\sum_{n=0}^\infty \frac{x^n}{n!}

You mean choosing 0 out of all possible real numbers? Then see this: http://en.wikipedia.org/wiki/Probability_distribution#Continuous_probability_distribution

Number of ways for none of Bob's numbers to be chosen = 34C6 (from the 34 numbers which Bob didn't choose, choose 6 of them) Total number of ways of choosing 6 numbers from 40 = 40C6 Hence Probability = Favourable/Total = 34C6/40C6

Not necessarily... the other case is that x=0 is also a solution to (x-3)² = k, which implies k=9. k=0 works as well obviously.

x is in radians so: x^\circ = \frac{\pi x}{180} Then substitute so x cancels to get limit