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HSC 2013-14 MX1 Marathon (archive) (8 Viewers)

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Stygian

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Re: HSC 2013 3U Marathon Thread

Yea it is, my initial fact was useless

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Yep i was thinking it was a red herring, all you had to do was determine the number of multiples of 6 between 1000 and 9999
 
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Re: HSC 2013 3U Marathon Thread

Yea it is, my initial fact was useless

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dafaq? that's not 3U. I can't use latex but my answer is Pi/2 I rationalsed and integrated.



my question:

find all solutions for:

3cos(2a) - sin(a) +2 = 0
 

aDimitri

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Re: HSC 2013 3U Marathon Thread

from a past 3u trial paper


post images
does this mean only 1 replacement occurs, and from the second draw on they stay out of the bag? terribly ambiguous question. i'm going to assume you mean you replace them all.

i) (1/3)^2 * (2/3)^2 * 4C2 = 24/81
ii) 4 are drawn, for the sum to be greater than 9, need 2 or more white counters.
P = 24/81 + 8/81 + 1/81 = 33/81
 
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Kurosaki

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Re: HSC 2013 3U Marathon Thread

does this mean only 1 replacement occurs, and from the second draw on they stay out of the bag? terribly ambiguous question. i'm going to assume you mean you replace them all.

i) (1/3)^2 * (2/3)^2 = 4/81
ii) 4 are drawn, for the sum to be greater than 9, need 2 or more white counters.
P = 4/81 + 2/81 + 1/81 = 7/81
Think you need to multiply by for the first bit to account for ordering.
Same for second bit (haven't done the problem myself though, just glancing through so I might be wrong).
 

mreditor16

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Re: HSC 2013 3U Marathon Thread

Kurosaki! done with trials?
 

aDimitri

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Re: HSC 2013 3U Marathon Thread

Think you need to multiply by for the first bit to account for ordering.
Same for second bit (haven't done the problem myself though, just glancing through so I might be wrong).
oh true that. didn't account for ordering hold up let me fix
 

Stygian

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Re: HSC 2013 3U Marathon Thread

does this mean only 1 replacement occurs, and from the second draw on they stay out of the bag? terribly ambiguous question. i'm going to assume you mean you replace them all.

i) (1/3)^2 * (2/3)^2 = 4/81
ii) 4 are drawn, for the sum to be greater than 9, need 2 or more white counters.
P = 4/81 + 2/81 + 1/81 = 7/81
this is the more logical assumption so you would go with it imo

would you say it would be erroneous to consider for part ii (in a situation where you weren't given part i) to work with 1-(probability of 3 or more black counters)? Should this logically give the same result?
 

Stygian

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Re: HSC 2013 3U Marathon Thread

Think you need to multiply by for the first bit to account for ordering.
Same for second bit (haven't done the problem myself though, just glancing through so I might be wrong).
^this should fix that
 

Kurosaki

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Re: HSC 2013 3U Marathon Thread

Kurosaki! done with trials?
Almost, I have 2 next week and then I'm done, but I have the rest of this week off haha :D.

Pretty nice question in my 3U trial that I'd like to share:

PQRS is a cyclic quadrilateral. Angle QPS is 90 degrees, angle PSR is , PQ=a, PS=b, QR=c.
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aDimitri

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Re: HSC 2013 3U Marathon Thread

Almost, I have 2 next week and then I'm done, but I have the rest of this week off haha :D.

Pretty nice question in my 3U trial that I'd like to share:

PQRS is a cyclic quadrilateral. Angle QPS is 90 degrees, angle PSR is , PQ=a, PS=b, QR=c.
Show
spent 10 minutes trying to figure this out then noticed the question said CYCLIC.
fml
 

mreditor16

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Re: HSC 2013 3U Marathon Thread

Almost, I have 2 next week and then I'm done, but I have the rest of this week off haha :D.

Pretty nice question in my 3U trial that I'd like to share:

PQRS is a cyclic quadrilateral. Angle QPS is 90 degrees, angle PSR is , PQ=a, PS=b, QR=c.
Show
ah nice, good luck with the remaining ones! :D
 

aDimitri

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Re: HSC 2013 3U Marathon Thread

Almost, I have 2 next week and then I'm done, but I have the rest of this week off haha :D.

Pretty nice question in my 3U trial that I'd like to share:

PQRS is a cyclic quadrilateral. Angle QPS is 90 degrees, angle PSR is , PQ=a, PS=b, QR=c.
Show
 

Sy123

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Re: HSC 2013 3U Marathon Thread

 
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