CSSA MX2 Trial (2 Viewers)

[RealiseNothing]

did you just count it or can you show working?.

Consider that you definately choose N, G, and E. The last two spots are filled with either R or A - RR, AA, AR. So you have 3 ways of choosing this.

Consider you only choose two of the three letters N, G, and E. The last three spots are filled with either R or A - RRA, RAA. So you have 2 ways of choosing it, but times by 3C2 since you are only choosing two out of the three letters. 2x(3C2)=6.

Consider you only choose one of the three letters N, G, and E. The last four spots are taken up by R or A - AARR. There is only one way of choosing this, but times by 3C1 since you are only choosing one out of the three letters. 1x(3C1)=3.

Hence add up all possible cases: 3+6+3 = 12.

 

[khfreakau]

agreed that answer is 12.

 

[maths lover]

Consider that you definately choose N, G, and E. The last two spots are filled with either R or A - RR, AA, AR. So you have 3 ways of choosing this.

Consider you only choose two of the three letters N, G, and E. The last three spots are filled with either R or A - RRA, RAA. So you have 2 ways of choosing it, but times by 3C2 since you are only choosing two out of the three letters. 2x(3C2)=6.

Consider you only choose one of the three letters N, G, and E. The last four spots are taken up by R or A - AARR. There is only one way of choosing this, but times by 3C1 since you are only choosing one out of the three letters. 1x(3C1)=3.

Hence add up all possible cases: 3+6+3 = 12.

pretty hard for a multiple choice question where you have no mark allocation to determine the difficulty of the question.

 

[RealiseNothing]

pretty hard for a multiple choice question where you have no mark allocation to determine the difficulty of the question.

There could be an easier way to do it using formulas, but I'd rather stay away from them in probability.

 

[khfreakau]

pretty hard for a multiple choice question where you have no mark allocation to determine the difficulty of the question.

i disagree, for a 4u probability question that is exceedingly easy. it's a typical question where you get cases, and it's not like they asked you to determine how many ways it can be arranged, which is a bit harder, but definitely not as hard as it can get. it could have been much harder, for sure.

 

[umm what]

Was it hard or easy?

 

[barbernator]

i disagree, for a 4u probability question that is exceedingly easy. it's a typical question where you get cases, and it's not like they asked you to determine how many ways it can be arranged, which is a bit harder, but definitely not as hard as it can get. it could have been much harder, for sure.

I agree with this freak. get it . yeh it is very easy. even in a minute you could write down all the selections without doing any maths!

 

[Trebla]

from memory:
16b)
i) 1 mark algebraic proof --> super easy
ii) recurrence formula for 3 marks using (i) identity... something like prove I(2n+1) = (2n/(2n+1)) x I(2n-1)
iii) prove I(2n+1) = (2^n * n!)/(1x3x5x...x(2n-1)) (2/3marks, can't remember)
iv) i couldn't even get close to typing this one out. it basically was a proof using a summation of iii) for a number of different values of n to get sigma (1/(2n+1)^2) (2marks)
v) using arcsin = x + uglything, prove 1 + 1/3^2 + 1/5^2 ... = pi^2 / 8 (2marks)
vi) find 1 + 1/2^2 + 1/3^2 + ... (2marks

Holy shit, I had effectively the same question in a trial paper I wrote for my students last year. Carrotsticks should know what I'm talking about cos he has a copy of it haha.

 

[Carrotsticks]

Proofs for famous identities and ESPECIALLY infinite series tend to be popular Q6/8 because the HSC is primarily Calculus-based, and a proper attempt at Analysis questions requires a strong understanding of just about everything.

 

[U MAD BRO]

Proofs for famous identities and ESPECIALLY infinite series tend to be popular Q6/8 because the HSC is primarily Calculus-based, and a proper attempt at Analysis questions requires a strong understanding of just about everything.

haha carrot loves series =]

 

[maths lover]

I agree with this freak. get it . yeh it is very easy. even in a minute you could write down all the selections without doing any maths!

i understand that it couldhave been done quite easily but it seems quite hard to do it using explanations etc, however probability is perhaps my weakest point anyways.

 

[bloodvial]

Have people started getting marks back yet?

Got mine back today, 71% and rank 1/3 (a pleasant surprise, the other two are crazy good usually). I'm pretty happy with that.

 

[OMGITzJustin]

Have people started getting marks back yet?

Got mine back today, 71% and rank 1/3 (a pleasant surprise, the other two are crazy good usually). I'm pretty happy with that.

how did the others go?

 

[bloodvial]

how did the others go?

Very close together, 2nd was 70% and 3rd 67%

 

[OMGITzJustin]

Very close together, 2nd was 70% and 3rd 67%

niiiiiice, still nervously waiting on my trial

 

[Carrotsticks]

Saw the questions for the MX2 CSSA.

The last question was pretty nice, except not quite happy about the fact that to complete one of the questions, you have to swap an integral and a summation sign under the assumption that it's allowed.

At least say something like "You may freely interchange a summation and the integral, without proof".

But overall this is something I would expect to be HSC difficulty, maybe a wee bit more difficult for the last question.

 

[asianese]

When does the security period end? ...I wanna see it

 

[RealiseNothing]

Saw the questions for the MX2 CSSA.

The last question was pretty nice, except not quite happy about the fact that to complete one of the questions, you have to swap an integral and a summation sign under the assumption that it's allowed.

At least say something like "You may freely interchange a summation and the integral, without proof".

But overall this is something I would expect to be HSC difficulty, maybe a wee bit more difficult for the last question.

The last question wasn't that hard in my opinion, I managed to get it out before you wrote the solution up.

 

[Aysce]

When does the security period end? ...I wanna see it

It already ended.

 

[Carrotsticks]

The last question wasn't that hard in my opinion, I managed to get it out before you wrote the solution up.

Exactly. Lots of bark, little bite.