HSC 2013-14 MX1 Marathon (archive) (1 Viewer)

Carrotsticks · Oct 8, 2013

Re: HSC 2013 3U Marathon Thread

Sy123 said:
The answer is 12 apparently lol

12 is correct.

JJ345 · Oct 8, 2013

Re: HSC 2013 3U Marathon Thread

VBN2470 said:
Please explain further...

I think he meant 12 is correct for the latter problem with the men and women alternating, not for the original one you were solving. I'm not sure.

mahmoudali · Oct 8, 2013

Re: HSC 2013 3U Marathon Thread

$i) Find the probability that in$ \ \ (n+k) \ \ $tosses of a fair coin, there will be$ \ \ n \ \ $heads$

$ii) Hence explain why$

$\sum_{k=0}^n \frac{\binom{n+k}{n}}{2^{n+k}} = 1$

Sy123 · Oct 8, 2013

Re: HSC 2013 3U Marathon Thread

Carrotsticks said:
I don't understand the bolded part. Can you clarify?

The problem is very difficult in its generalised form. I can imagine that it would use some form of the inclusion/exclusion principle.

Well, essentially we take an arrangement of 3 couples that are already arranged so that no one sits with their couple. Then we add in another 2 people,

(imagine this line to be a circle)

So we have

A _ B _ A _ C _ B _ C _

That is an arrangement of 3 couples such that no couple is together, the _'s indicate spaces where we can put in the 2 couples DD.
Therefore, for every arrangement of 3 couples, there are 6C2 ways for the next couple to sit to make 4 couples arranged in a desired fashion.

i.e. $u_4 = C(6,2) u_3$

Is there something wrong with this?

Carrotsticks said:
12 is correct.

That was meant for the same problem but the couples were alternating, I copied the question down incorrectly.

JJ345 · Oct 8, 2013

Re: HSC 2013 3U Marathon Thread

I think you've missed cases like this
3 couples initially: A_A_B_C_B_C
Where one D comes and fills the first marker and the second D can choose from the others.

Sy123 · Oct 8, 2013

Re: HSC 2013 3U Marathon Thread

JJ345 said:
I think you've missed cases like this
3 couples initially: A_A_B_C_B_C
Where one D comes and fills the first marker and the second D can choose from the others.

Ahh yes that is true, thank you

BagOFyarn · Oct 15, 2013

Re: HSC 2013 3U Marathon Thread

mahmoudali said:
$i) Find the probability that in$ \ \ (n+k) \ \ $tosses of a fair coin, there will be$ \ \ n \ \ $heads$

$ii) Hence explain why$

$\sum_{k=0}^n \frac{\binom{n+k}{n}}{2^{n+k}} = 1$

ColdLipstick · Oct 17, 2013

Re: HSC 2013 3U Marathon Thread

Sy123 · Oct 18, 2013

Re: HSC 2013 3U Marathon Thread

$Find$

$\sum_{k=0}^m \binom{j}{k} \binom{i}{m-k}$

HeroicPandas · Oct 18, 2013

Re: HSC 2013 3U Marathon Thread

Sy123 said:
$Find$

$\sum_{k=0}^m \binom{j}{k} \binom{i}{m-k}$

Consider $(1+x)^i (1+x)^j = (1+x)^{i+j}$

Equate co-eff of x^m

So.... $\sum_{k=0}^m \binom{j}{k} \binom{i}{m-k} = \binom{i+j}{m}$ ?

Sy123 · Oct 18, 2013

Re: HSC 2013 3U Marathon Thread

HeroicPandas said:
Consider $(1+x)^i (1+x)^j = (1+x)^{i+j}$

Equate co-eff of x^m

So.... $\sum_{k=0}^m \binom{j}{k} \binom{i}{m-k} = \binom{i+j}{m}$ ?

Yep

Sy123 · Oct 19, 2013

Re: HSC 2013 3U Marathon Thread

$Prove by mathematical induction that the sum of interior angles of an$ \ n \ $sided polygon is$

$180(n-2) \ \ $in degrees measure$$

Drewk · Oct 19, 2013

Re: HSC 2013 3U Marathon Thread

http://snag.gy/pLUT9.jpg
Trying this past hsc question using 1/2v^2
because we r given v i squared both sides then got 2xe^-x^4
then i would differentiate and mul by 1/2 but i don't think that gets me to the answer
Please help

ColdLipstick · Oct 19, 2013

Re: HSC 2013 3U Marathon Thread

Drewk said:
http://snag.gy/pLUT9.jpg
Trying this past hsc question using 1/2v^2
because we r given v i squared both sides then got 2xe^-x^4
then i would differentiate and mul by 1/2 but i don't think that gets me to the answer
Please help

Your method is correct.

HeroicPandas · Oct 20, 2013

Re: HSC 2013 3U Marathon Thread

$$ A goat grazes a rectangular paddock 10 metres by 20 metres. It is tethered to the fence at one corner of the paddock by an inextensible rope of length x metres, where 10 $< x < 20$. \\ \\ $ $Show that the goat can graze an area of $A m^2$ where \\ \equation{$ A = \frac{1}{2}x^2 sin^{-1}\left ( \frac{10}{x} \right ) + 5 \sqrt{x^2 -100}}$

asianese · Oct 20, 2013

Re: HSC 2013 3U Marathon Thread

ColdLipstick said:
Your method is correct.

Use $\dfrac{d}{dx}$ rather than $\dfrac{\delta}{\delta x}$ . Deltas have a different implication and aren't use in the syllabus.

dunjaaa · Oct 20, 2013

Re: HSC 2013 3U Marathon Thread

HeroicPandas said:
$$ A goat grazes a rectangular paddock 10 metres by 20 metres. It is tethered to the fence at one corner of the paddock by an inextensible rope of length x metres, where 10 $< x < 20$. \\ \\ $ $Show that the goat can graze an area of $A m^2$ where \\ \equation{$ A = \frac{1}{2}x^2 sin^{-1}\left ( \frac{10}{x} \right ) + 5 \sqrt{x^2 -100}}$

ill just write up the solution, latex fail...

dunjaaa · Oct 20, 2013

Re: HSC 2013 3U Marathon Thread

Heres my solution to ^

HeroicPandas · Oct 20, 2013

Re: HSC 2013 3U Marathon Thread

dunjaaa said:
View attachment 28843 Heres my solution to ^

Very nice

phoenix159 · Oct 21, 2013

Re: HSC 2013 3U Marathon Thread

Integration - Volume of Solids of Revolution

A hemispherical bowl of radius a units is filled with water to a depth of a/2 units. Find, by integration, the volume of the water.

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