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HSC 2015 MX1 Marathon (archive) (1 Viewer)

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rand_althor

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Re: HSC 2015 3U Marathon



1. (a) 2/75075 (b) 11/13
2. (a) 1/11550 (b) 1/55
 

davidgoes4wce

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Re: HSC 2015 3U Marathon

Having a bit of trouble on this one, I understand how to get the derivative but isn't the derivative at x=0, f'(x)=0 (which in this case is neither decreasing not increasing)?

 

Drsoccerball

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Re: HSC 2015 3U Marathon

Having a bit of trouble on this one, I understand how to get the derivative but isn't the derivative at x=0, f'(x)=0 (which in this case is neither decreasing not increasing)?

proving that its a decreasing function means that f(x)<0
 

davidgoes4wce

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Re: HSC 2015 3U Marathon

Could you also use concavity to explain that as well? This was what another friend of mine told me today. You can use the 2nd derivative to prove that its a decreasing function.
 

InteGrand

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Re: HSC 2015 3U Marathon

Could you also use concavity to explain that as well? This was what another friend of mine told me today. You can use the 2nd derivative to prove that its a decreasing function.
No, because you can have increasing and decreasing functions with either concavity, e.g. and are both concave down, but one is increasing, whilst the other is decreasing. You can use the second derivative to prove something about the increasing or decreasing nature of the first derivative though, if that's helpful for a problem (which it sometimes is).
 

Henrybro69

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Re: HSC 2015 3U Marathon

(i) Express −3 sinx − 4 cos x in the form S cos(x − β), where S > 0 and 0 < β < 2π.
(Write β to four decimal places.)
(ii) Hence solve −3 sinx − 4 cos x = 2, for 0 ≤ x < 2π. Give the solutions correct to
two decimal places.
without radians in answer
 

davidgoes4wce

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Re: HSC 2015 3U Marathon

(i) Express −3 sinx − 4 cos x in the form S cos(x − β), where S > 0 and 0 < β < 2π.
(Write β to four decimal places.)
(ii) Hence solve −3 sinx − 4 cos x = 2, for 0 ≤ x < 2π. Give the solutions correct to
two decimal places.
without radians in answer

Part (i)

 

calamebe

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Re: HSC 2015 3U Marathon

Yeah, and once you have that formula you let -5cos(x-0.6435)=2, and manipulate it until you have cos(x-0.6435) on it's on on the left hand side, and use inverse cos to find what x-0.6435 is equal to, and then add 0.6435.

Edit: nevermind he posted after I started the message.
 

davidgoes4wce

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Re: HSC 2015 3U Marathon

I am not a poker player but this question came up.

A hand of 5 cards is dealt out from a well-shuffled pack of 52 cards. Find the probability there is 'a straight flush'.
Having a quick read about what a 'flush ' was 'A straight flush is a hand that contains five cards in sequence, all of the same suit'

my possible combinations was:
{A,2,3,4,5}
{2,3,4,5,6}
{3,4,5,6,7}
{4,5,6,7,8}
{5,6,7,8,9}
{6,7,8,9,10}
{10,J,Q,K,A}

I understand that if you then multiply each of these by the 4 suits , you obtain 4x7=28.

My question is why do they go for 4 x 10? Which combination of cards am I missing?
 

Drsoccerball

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Re: HSC 2015 3U Marathon

I am not a poker player but this question came up.

A hand of 5 cards is dealt out from a well-shuffled pack of 52 cards. Find the probability there is 'a straight flush'.
Having a quick read about what a 'flush ' was 'A straight flush is a hand that contains five cards in sequence, all of the same suit'

my possible combinations was:
{A,2,3,4,5}
{2,3,4,5,6}
{3,4,5,6,7}
{4,5,6,7,8}
{5,6,7,8,9}
{6,7,8,9,10}
{10,J,Q,K,A}

I understand that if you then multiply each of these by the 4 suits , you obtain 4x7=28.

My question is why do they go for 4 x 10? Which combination of cards am I missing?
The possible amount of cards you pick such that the minimum number is the min number of the straight flush is (8C1) but you have to also choose a suit so you multiply by (4C1) so then you divide by the total sample space being (52C5)
therefore your answer should be:

EDIT: I think you got it wrong because you didnt know the rules
First of all 7,8,9,10,J 8,9,10,J,Q 9,10,J,Q,Kare also combinations
In my solution i didnt include the ace as a term so it should b (9C1) rather than 8C1
 
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