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No... just no.

Clearly they have never heard of scaling before

^

Scaling has nothing to do with the difficulty of a subject, but rather the performance of the subject's cohort in other exams. Although 'difficult' subjects do tend to scale better (since 'smarter' students are more likely to pick them anyway), correlation does not imply causation. E.g. Your...

I'm pretty sure 20 is D since $30 is the domestic equilibrium market price so there are no imports to be taxed at all Also shouldn't 18 be D since an increase in the supply of oil prices leads to lower prices and hence lower headline inflation? I stuffed up the superannuation MC though :(

I cringed when I saw the two 4 and 6 markers on environmental sustainability. Thank god I had enough sense to immediately skip it and come back to it later when my bullshitting skills improved during the exam

Y = C + I but C= 50 + 0.8Y therefore Y= 50 + 0.8Y + 100 then solve

Re: HSC 2013 4U Marathon Or just a 'HSC 4U Marathon' thread, let it last indefinitely :p

Re: HSC 2013 4U Marathon Is it possible to change the thread title?

You're confusing aligning with scaling

combinomialations seem to make the most sense, don't think that should count lol

I can't believe I couldn't get q10 out when the solution is pretty standard. Amazing what exam pressure/stress can do to you :(

I just worked q10 out I'm pretty sure it's actually A total number of ways = 4096 - 4 (number of ways all 6 people can in be one room) - 6 x 4 (number of ways 5 people can be chosen out of 6 to be in one room) x 3 (leftover person choosing the rest of the rooms) = 4020 Of course I have to...

What do you guys think the E4 cut off will be?

Never thought I'll ever see a question that involves both conics and mechanics...

Just edited my post. I guess I shouldn't be too extreme based on one person's experience haha

After witnessing my brother doing pharmacy for the past few years, I'm afraid OP has some element of truth.

Re: MX2 Integration Marathon \\I=\int_{0}^{n}xe^{-ax}dx\\\\=\left [ -\frac{1}{a}xe^{-ax}\right ]_{0}^{n}+\frac{1}{a}\int_{0}^{n}e^{-ax}dx\\\\=-\frac{n}{a}e^{-an}+\frac{1}{a}\left [-\frac{ e^{-ax}}{a} \right ]_{0}^{n}\\\\=-\frac{n}{a}e^{-an}-\frac{1}{a^2}e^{-an}+\frac{1}{a^2}\\\\\therefore...

Yeah but it still went from 1 to 39 (or however many people took the test)

Yeah, I think he's trying to make us feel better after all we've been through :p