VCE Methods Question Thread!

[TheAspiringDoc]

Could someone mind helping me out?

I got A but apparently the answer is meant to be B.

Any ideas why?

What were your workings?
I'm just not sure where to start, other than drawing a graph..

[TheAspiringDoc]

If a piece of wire is 80cm long find the area of the largest rectangle that can be formed by the wire.

Isn't it just 400cm^3?
Like kinda just an intuitive thing..

[keltingmeith]

Isn't it just 400cm^3?
Like kinda just an intuitive thing..

Why would it be that? Give me your reasoning. Remember, this is a maths question, so I want a proper mathematical argument.

[keltingmeith]

Hi guys

I was doing the Kilbaha 2004 exam when I came across something called, "Hypergeometric" formulas. I haven't seen one in my cambridge textbook. Is this in the methods course?

One thing that I noticed, was that I was able to solve the question much more quickly using that formula rather than having to do tedious calculations.

Could someone explain this for me? Is this safe to use in the methods exam (a way to double check answers or something)

Thank you!

Hmm, not sure what this "Hypergeometric" formula is, unless you're referring to the hypergeometric distribution. Reckon you could share it for us?

I will note that the current study design started in 2006, so a 2004 exam (particularly from Kilbaha) could stray from the current study design.

[TheAspiringDoc]

Why would it be that? Give me your reasoning. Remember, this is a maths question, so I want a proper mathematical argument.

True.
I guess it's just that I know for a fact that squares have the largest area : total side length ratio for any for sided polygon.
And if the the total side length is 80cm, and a square has four sides, then each side length would be 20cm.

[keltingmeith]

True.
I guess it's just that I know for a fact that squares have the largest area : total side length ratio for any for sided polygon.
And if the the total side length is 80cm, and a square has four sides, then each side length would be 20cm.

You're right - however, this is not a fact, it's a result. Definitely something you should consider trying to prove - this is a year 11 proof, IMHO. Maybe year 12 standard.

[TheAspiringDoc]

You're right - however, this is not a fact, it's a result. Definitely something you should consider trying to prove - this is a year 11 proof, IMHO. Maybe year 12 standard.

And proof by examples doesn't count does it?
I mean, you can't just say '5x5=25, 4x6=24. 8x8=64, 9x7=63. etc.'

[keltingmeith]

And proof by examples doesn't count does it?
I mean, you can't just say '5x5=25, 4x6=24. 8x8=64, 9x7=63. etc.'

Nope. But examples might help you think about what way to do the proof.

[StupidProdigy]

Could someone mind helping me out?

I got A but apparently the answer is meant to be B.

Any ideas why?

A is correct

[StupidProdigy]

Got any nice links to a proof for it, I'm interested 

[TheAspiringDoc]

Got any nice links to a proof for it, I'm interested 

Is this what you were thinking of?
https://purplemath.com/modules/perimetr6.htm

[Apink!]

Hmm, not sure what this "Hypergeometric" formula is, unless you're referring to the hypergeometric distribution. Reckon you could share it for us?

I will note that the current study design started in 2006, so a 2004 exam (particularly from Kilbaha) could stray from the current study design.

Here you go!

[keltingmeith]

Here you go!

Okay, so it is the hypergeometric distribution.

I agree that the distribution is easier and nicer to work with than doing the counting yourself, BUT I would advise against its use as I doubt many teachers are familiar with it.

I wouldn't worry, though - you should only get simple "draws without replacement" questions, anyway. Stuff you can do with tree diagrams.

[Apink!]

Thank you!

Does anyone know how I can graph binomial distribution graph on the CAS?

[HopefulLawStudent]

Can anyone help w 3a, please?