HSC 2013 Maths Marathon (archive) (3 Viewers)

spanky125 · Oct 21, 2013

Re: HSC 2013 2U Marathon

A company that manufactures anal beads has a 2% chance of producing a defective anal bead. If it produces 8000 anal beads an hour for 12 hours a day. How many of defective anal beads will be produced in five working days?

Drewk · Oct 21, 2013

Re: HSC 2013 2U Marathon

spanky125 said:
a company that manufactures anal beads has a 2% chance of producing a defective anal bead. If it produces 8000 anal beads an hour for 12 hours a day. How many of defective anal beads will be produced in five working days?

9600? Hahaha

Youi_ · Oct 21, 2013

Re: HSC 2013 2U Marathon

spanky125 said:
A company that manufactures anal beads has a 2% chance of producing a defective anal bead. If it produces 8000 anal beads an hour for 12 hours a day. How many of defective anal beads will be produced in five working days?

Total beads produced over 5 working days = 480,000

2% of 480,000 = 9,600 defective beads?

spanky125 · Oct 21, 2013

Re: HSC 2013 2U Marathon

Drewk said:
9600? Hahaha

Yes 9600

Youi_ · Oct 21, 2013

Re: HSC 2013 2U Marathon

Parramatta Eels have a 0.1 chance of winning a game they play. How many games must be played so that they are 99% certain of winning.

Triage · Oct 21, 2013

Re: HSC 2013 2U Marathon

Drewk said:
why do you use "7^x = e^(xln7)" ??

Because you can integrate and differentiate that. It's in the course somewhere

spanky125 · Oct 21, 2013

Re: HSC 2013 2U Marathon

Youi_ said:
Parramatta Eels have a 0.1 chance of winning any game they play. How many games must be played so that they are 99% certain of winning.

99% certain? It doesn't matter how many games they play, if they have a 0.1 chance of winning any game they play then they will never get to 99%. From what you are saying previous results make no difference to the next games probability, therefore it will always be a 0.1 chance of winning.

Youi_ · Oct 21, 2013

Re: HSC 2013 2U Marathon

spanky125 said:
99% certain? It doesn't matter how many games they play, if they have a 0.1 chance of winning any game they play then they will never get to 99%. From what you are saying previous results make no difference to the next games probability, therefore it will always be a 0.1 chance of winning.

edited

Triage · Oct 21, 2013

Re: HSC 2013 2U Marathon

spanky125 said:
99% certain? It doesn't matter how many games they play, if they have a 0.1 chance of winning any game they play then they will never get to 99%. From what you are saying previous results make no difference to the next games probability, therefore it will always be a 0.1 chance of winning.

I'm pretty sure he/she means how many games do they have to play for their to be 99% chance they have won one of them

ocatal · Oct 21, 2013

Re: HSC 2013 2U Marathon

$$By considering triangle PBA, show that$ \ \sin\alpha = \frac{x^2 - 16}{4x}$

Youi_ · Oct 21, 2013

Re: HSC 2013 2U Marathon

Triage said:
I'm pretty sure he/she means how many games do they have to play for their to be 99% chance they have won one of them

He, and yes something along those lines. English HSC was last week I don't care much anymore about my grammatical skills until next year lol.

Triage · Oct 21, 2013

Re: HSC 2013 2U Marathon

ocatal said:
$$By considering triangle PBA, show that$ \ \sin\alpha = \frac{x^2 - 16}{4x}$

I tried using the cosine rule for cos(90-alpha) in triangle apb but I'm getting (x^2 - 6)/4x

ocatal · Oct 21, 2013

Re: HSC 2013 2U Marathon

Triage said:
I tried using the cosine rule for cos(90-alpha) in triangle apb but I'm getting (x^2 - 6)/4x

That's correct. Are you sure you're applying the cosine rule correctly?

Youi_ · Oct 21, 2013

Re: HSC 2013 2U Marathon

ocatal said:
$$By considering triangle PBA, show that$ \ \sin\alpha = \frac{x^2 - 16}{4x}$

Let alpha=a for simplicity

$cos(90-a) = sin(a)$

$(2\sqrt{5})^2=x^2 + 4 - 4xcos(90-a)$

$20= x^2 + 4 -4sin(a)$

$4sin(a) = x^2 - 16$

$sin(a) = \frac{x^2 -16}{4x}$

Triage · Oct 21, 2013

Re: HSC 2013 2U Marathon

ocatal said:
That's correct. Are you sure you're applying the cosine rule correctly?

Shit sorry. My calculator was giving me 2root5 squared as 10. My bad

ocatal · Oct 21, 2013

Re: HSC 2013 2U Marathon

Triage said:
Reasonably

Oh I realise what you've done. When squaring $2\sqrt{5}$ you must also square the 2 in the front. So you would get 4 x 5 = 20.

Triage · Oct 21, 2013

Re: HSC 2013 2U Marathon

ocatal said:
Oh I realise what you've done. When squaring $2\sqrt{5}$ you must also square the 2 in the front. So you would get 4 x 5 = 20.

Yeah exactly.

ocatal · Oct 21, 2013

Re: HSC 2013 2U Marathon

Triage said:
Yeah exactly.

Not sure why you aren't getting x^2 -16 then.

Triage · Oct 21, 2013

Re: HSC 2013 2U Marathon

ocatal said:
Not sure why you aren't getting x^2 -16 then.

Oh no I've got it now. My calculator was squaring it incorrectly

Youi_ · Oct 21, 2013

Re: HSC 2013 2U Marathon

Triage said:
Oh no I've got it now. My calculator was squaring it incorrectly

Yes blame it on the infallible calculator <> hahaha my calculator sometimes loses its mind too.

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