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    MATH1131 vs MATH1141

    Do 1141, scaling is epic compared to 1131 and its not much harder.
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    Easiest finance elective

    All finance courses are WAM boosters ... just pick any
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    permutation help!

    1) Since you want the letters of KOALA in order, fix those first: __ K __ O __ A __ L __ A __ The lines indicate the spots the letters of GUMTREE can be placed 2) Choose the spot for the letter G: 6 spots means 6 ways to do so. Then our arrangement becomes something like (depending on...
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    permutation and combination

    Number of ways with both students with at least 1 book = Total number of ways - Number of ways that one student has no books Using that, we have: Total number of ways= 2x2x2x2x2x2 = 2^6 (each book can go to 2 positions) Number of ways where one student has no books =2 Hence Number of ways...
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    help, with polynomial question!!

    This is what Carrotsticks is referring to:
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    Probability Question, need help with a logial way.

    What's the probability of rolling a 2 and a 1? Compare this with the probability of rolling two 1s. That should answer your questions.
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    Commerce queries

    For IBs its more like 80 WAM
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    Finance Electives

    I did SAV and IRRM, both of which are pretty easy. None of them really that interesting to be honest. It doesn't really matter which finance course you choose as they are generally good WAM boosters. Henry Yip's tests aren't hard and he scales pretty generously too.
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    Higher vs Normal (Maths)

    normal generally have slightly less content (around 1 topic less than higher). For higher, the topics they share with the normal level would include an extension. This is compensated by the fact is much easier to get a higher mark (HD+) in higher than normal if you are decent at maths.
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    MATH1081 and MATH1231

    If you like maths, do 1081 cos its a lot more interesting than 1231 (which will be mandatory anyway). Otherwise its not necessary.
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    Fun little problem

    just work backwards, screw using equations and algebra
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    another complex question

    Although to do it properly is difficult, using some tricks make it easy. These tricks shouldn't really be used for normal questions. This is really a more educated trial and error keeping in mind the design of the question. The idea is to note that since we are adding 2 real numbers on left...
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    worth buying Matlab?

    No you don't use matlab a lot in any commerce degree. Seriously though, why would you buy it anyway. There are plenty of ways around it.
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    Complex number

    See that \frac{z_1}{z_2}=r(\cos\theta+i\sin\theta) where r is the modulus and \theta is the argument. Your aim is then to find the theta and r. r is obvious to find. Consider \theta. We know that the sides z1, z2, z1-z2 forms a triangle with side lengths 3,4 and \sqrt{37} respectively (draw a...
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    HSC 2012 MX2 Marathon (archive)

    Re: 2012 HSC MX2 Marathon question kinda hints to use GP Noting that |\sin x|< 1 for 0 < x < \frac{\pi}{2}, see that: \int \frac{\sin x+1}{\cos x} dx &= \int \frac{1-\sin^2 x}{\cos x (1-\sin x)} dx = \int \frac{\cos x}{1-\sin x} dx=\int \cos x \sum_{n=0}^\infty \sin^n x dx Assuming...
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    No need for textbook. Lecture notes should be enough.
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    MATH1151 Asoc Prof Womersley and Tran - Actuarial Studies

    In MATH1151, several variable calculus takes a bit of time to get used to. Otherwise no topic is particularly hard to grasp. MATLAB is not hard at all (at least what you need to know). Do all the practice questions and you'll be fine.
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    discrete maths math1081 worth doing

    Why not do both? MATH1081 is quite useful when learning programming as it gives you a good grasp on logical flow. Plus learning how to construct rigorous arguments in proofs is still necessary in applied and stats as well and all the compulsory "pure" maths subjects in second year. MATH1081 is...
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    Need help on Probability Question!

    a) Suppose we have Car 1 Faulty, other 2 cars fine. Then probability is 3/100*97/100*97/100. Since Car 2 or Car 3 can also be faulty, there are 3 possible combinations in total, hence you multiply by 3. b) probability for 1 car to not be faulty =97/100, rest is self explanatory c) Prob of 1 car...
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    Need help on Probability Question!

    Noting that 1 car has 3/100 chance of being faulty... a) 3/100*97/100*97/100*3 b) (97/100)^3 c) (3/100)^3