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The probabilities for {1,2,3,4,5,6} are {x,x,x,x,x,2/3} where x is unknown. You can solve for x using the fact that sum of the probabilities across the entire sample space is 1. Once you do that, you will have the full probability distribution and can answer the rest of the questions.

By conservation of mass we must have that Q_{\text{tank}}=Q_{\text{initial}}+Q_{\text{in}}-Q_{\text{out}} However, since there is no salt coming into the tank (i.e. Q_{\text{in}}=0) the problem now reduces to figuring out how much salt is coming out the tank at time t, so...

Raw marks are aligned to give aligned marks. Aligned marks are NOT scaled any further and do not form any part of the ATAR calculation. Aligned marks purely exist for the HSC record of achievement, not the ATAR. Raw marks are scaled to give scaled marks. These are used in the ATAR calculation...

There is no material mathematical advantage for your ATAR either way. It all depends on what environment suits you to perform better, as it all comes from within yourself. For example, there is no point going to a higher ranked school if the thought of half the grade being smarter than you...

For future reference, please post any questions for specific subjects in the subject forums - not the general HSC forums. Thanks

Calm down children...

Prove by induction. In your inductive step, you will need to use the result k² > 2k+1 for integers k > 4 in order to retrieve the perfect square. This result can be proven in a number of ways. Perhaps the shortest approach is as follows: k > 4 k² > 4k k² > 2k + 2k > 2k + 1 (since 2k > 8 > 1)

The order of the DE is effectively the number of times you need to “integrate” it to get the general solution. Each time you integrate you generate an arbitrary constant which cannot be found unless you know the specific conditions.

If you assume a ground level projection then when you specify the coordinate axes you have three known points (0,0), (0.5R,H) and (R,0) where R is the range and H is the maximum height. Three points allows you to specify the exact Cartesian equation of the parabola that models the projectile...

I think the user meant when all parties have a choice. In a murder, one party doesn’t exactly get a choice...

Can you please post your questions in the Maths forum, NOT the secondary education forum I have had to move your thread.

Moved to Maths Ext1 thread. For future reference, please post these questions in the Maths forum.

The mean is based on the values of the data points. The median is based on the rank of the data points, rather than their value. The former is therefore sensitive to big outliers or extreme values which can distort the measure of the “centre”. The median does not care how big/small the extreme...

Decimal places would naturally exist when you apply weightings to percentages in a single assessment. I’m pretty sure schools submit marks with decimal places, though it is not clear how many decimal places.

The whole point of the subject is not intended to revolve around a single theory exam. The external HSC exam is only ONE way to assess syllabus outcomes. The subject is supposed to teach and assess a diverse range of skills that are applicable to uni/workplace etc, not just how to answer...

What you are suggesting would completely defeat the purpose of having other assessments lol School assessments exist as they can cover skills specified in the syllabus beyond just a theory exam. For example, you can’t claim to be good at Chemistry if you are hopeless at applying theory into...

There is not enough information because you haven’t specified how the rest of your cohort performs in both internal vs external. If we assume that the: - shape of the distribution of your cohort‘s internal and external marks are the same - there is no one between you and the other student both...

Are you only talking about swapping internal marks? If so, then that is trivial as you and the other person obviously just swap marks. If you’re talking about having internal marks and then swapping places on external marks then that’s a little different.

Half angle formula is literally just the double angle formula. The only other half angle thing is t-formula. Either way, you don’t really need to remember them as they are more or less derived from the “Compound angles” section of the formula sheet. If you can derive it from these formulae then...

Do it sequentially: - There are \binom{12}{6} ways to select 12 people to form the table of 6 - Once those 6 are selected there are 6 left so there are \binom{6}{4} ways to select 6 people to form the table of 4 - Once those 4 are selected there are 2 left so there is \binom{2}{2} way to form...