Author Topic: Mathematics Question Thread (Read 1626454 times) Tweet Share

georgiia · « **Reply #3015 on:** October 22, 2017, 04:11:09 pm »

Quote from: jamonwindeyer on October 22, 2017, 04:07:44 pm

Basically, it is because you are assigning a probability to which person you pick, when really that isn't something that matters - Because all of their opinions are counted.

The best way to think of it is to think, well, if all three are for it, then

Person A is for it
AND
Person B is for it
AND
Person C is for it

So you can turn that into an equation (which a tree diagram helps you visualise) -

$P=P\left(\text{ A Agrees}\right)\times P\left(\text{ B Agrees}\right)\times P\left(\text{ C Agrees}\right)$

Remember, the word AND means you multiply, the word OR means you add the respective probabilities

Ok thanks, now I’m struggling with the second part

RuiAce · « **Reply #3016 on:** October 22, 2017, 04:13:16 pm »

Quote from: georgiia on October 22, 2017, 04:11:09 pm

Ok thanks, now I’m struggling with the second part

$\text{If the majority is in favour we must have}\\ \mathbb{P}(FFF) + \mathbb{P}(FFA) + \mathbb{P}(FAF) + \mathbb{P}(AFF)$
$\text{So }0.7^3 + (0.7)^2(0.3)+(0.7)^2(0.3)+(0.7)^2(0.3)\\ \text{which indeed gives 0.784}$

jamonwindeyer · « **Reply #3017 on:** October 22, 2017, 04:14:37 pm »

Quote from: georgiia on October 22, 2017, 04:11:09 pm

Ok thanks, now I’m struggling with the second part

So the majority would mean 2 OR 3 people are supportive of the change! So, find the branches of your tree diagram that reflect this (there should be four of them) and add up their probabilities! Essentially you are doing:

- A, B AND C are all supportive (same as your answer from Part (i))
OR (meaning, plus)
- A AND B are supportive but C isn't
OR
- A AND C are supportive but B isn't
OR
- B AND C are supportive but A isn't

Which is why we take four branches

georgiia · « **Reply #3018 on:** October 22, 2017, 04:14:57 pm »

Quote from: RuiAce on October 22, 2017, 04:13:16 pm

$\text{If the majority is in favour we must have}\\ \mathbb{P}(FFF) + \mathbb{P}(FFA) + \mathbb{P}(FAF) + \mathbb{P}(AFF)$
$\text{So }0.7^3 + (0.7)^2(0.3)+(0.7)^2(0.3)+(0.7)^2(0.3)$

I get it now with this type of diagram ! Thanks

emmy_b · « **Reply #3019 on:** October 22, 2017, 04:34:13 pm »

Hi can someone make worked solutions for 2013 HSC Q6? Super stuck!

http://educationstandards.nsw.edu.au/wps/wcm/connect/2cdf8537-92dc-4328-ab39-74a1f1dd489f/maths-hsc-exam-2013.pdf?MOD=AJPERES&CACHEID=ROOTWORKSPACE-2cdf8537-92dc-4328-ab39-74a1f1dd489f-lGhNsb6

K9810 · « **Reply #3020 on:** October 22, 2017, 04:58:33 pm »

Can you please help me with this question:

A superannuation fund pays an interest rate of 8.75% per annum which compounds annually. Stephanie decides to invest $5000 in the fund at the beginning of each year, commencing on 1 January 2003.
What will be the value of Stephanie’s superannuation when she retires on 31 December 2023?

Thanks

~BK~ · « **Reply #3021 on:** October 22, 2017, 05:05:15 pm »

hey...
could you pls help me out with the attached q from the 2011 paper...
don't even know where to start

still kinda just staring at it blankly
thanks in advance

RuiAce · « **Reply #3022 on:** October 22, 2017, 05:22:23 pm »

Quote from: emmy_b on October 22, 2017, 04:34:13 pm

Hi can someone make worked solutions for 2013 HSC Q6? Super stuck!

http://educationstandards.nsw.edu.au/wps/wcm/connect/2cdf8537-92dc-4328-ab39-74a1f1dd489f/maths-hsc-exam-2013.pdf?MOD=AJPERES&CACHEID=ROOTWORKSPACE-2cdf8537-92dc-4328-ab39-74a1f1dd489f-lGhNsb6

Back when I did this paper, I just subbed in points and looked out for which one worked.
$\text{But as for the actual method}\\ \text{we need to rewrite }\boxed{y=\sin \left[2\left(x+\frac\pi6\right)\right]}$
$\text{The phase shift is actually }-\frac{\pi}{6}\\ \text{We always need to factorise before writing down the phase.}$
The only one that intercepts the x-axis at -π/6 is D.

MisterNeo · « **Reply #3023 on:** October 22, 2017, 05:26:32 pm »

Quote from: ~BK~ on October 22, 2017, 05:05:15 pm

hey...
could you pls help me out with the attached q from the 2011 paper...
don't even know where to start still kinda just staring at it blankly
thanks in advance

$\text{You would use the distance formula}\\40=(\sqrt{(y-0)^2+(x+1)^2})^2+(\sqrt{(y-0)^2+(x-3)^2})^2 \\ 40=y^2+(x+1)^2+y^2+(x-3)^2 \\ 40=2y^2+x^2+2x+1+x^2-6x+9 \\ 40=2y^2+2x^2-4x+10$ $20=y^2+x^2-2x+5 \\ 16=(x-1)^2+y^2$ And you can find the centre/radius from this equation (1,0) r=4.

jamonwindeyer · « **Reply #3024 on:** October 22, 2017, 05:27:35 pm »

Quote from: K9810 on October 22, 2017, 04:58:33 pm

Can you please help me with this question:

A superannuation fund pays an interest rate of 8.75% per annum which compounds annually. Stephanie decides to invest $5000 in the fund at the beginning of each year, commencing on 1 January 2003.
What will be the value of Stephanie’s superannuation when she retires on 31 December 2023?

Thanks

Hey! What was your approach to starting the question? Did you have a go at forming the required series? So something like this would be your approach:

$A_0=5000\\A_1=5000(1.0875)+5000\\A_2=A_1(1.0875)+5000$

~BK~ · « **Reply #3025 on:** October 22, 2017, 05:34:40 pm »

Quote from: MisterNeo on October 22, 2017, 05:26:32 pm

$40=(\sqrt{(y-0)^2+(x+1)^2})^2+(\sqrt{(y-0)^2+(x-3)^2})^2 \\ 40=y^2+(x+1)^2+y^2+(x-3)^2 \\ 40=2y^2+x^2+2x+1+x^2-6x+9 \\ 40=2y^2+2x^2-4x+10$ $20=y^2+x^2-2x+5 \\ 16=(x-1)^2+y^2$ And you can find the centre/radius from this equation (1,0) r=4.

wow, that's awesome thanks so much

makes heaps more sense now, it just seemed to be worded strangely (as BOSTES love to do with any maths q)

Natasha.97 · « **Reply #3026 on:** October 22, 2017, 05:34:56 pm »

Quote from: ~BK~ on October 22, 2017, 05:05:15 pm

hey...
could you pls help me out with the attached q from the 2011 paper...
don't even know where to start still kinda just staring at it blankly
thanks in advance

Hi!

Hope this helps

_____ · « **Reply #3027 on:** October 22, 2017, 06:18:28 pm »

For this (iii):

Can I just sub in y into the equation given or do I need to "show" how the equation is found by 'going the other way' if you get what I mean? If so, how would this be done?

Thanks!

blasonduo · « **Reply #3028 on:** October 22, 2017, 06:21:08 pm »

Hey!
(I totally didn't accidentally put this in 4u

)

For questions like this (ii) how do I adequately prove that the value is always positive?

I have lost a mark in trials for not doing this, but I'm unsure how to do this apart from subbing in values and going "look its still positive!"

Natasha.97 · « **Reply #3029 on:** October 22, 2017, 06:26:14 pm »

Quote from: blasonduo on October 22, 2017, 06:21:08 pm

Hey!
(I totally didn't accidentally put this in 4u )

For questions like this (ii) how do I adequately prove that the value is always positive?

I have lost a mark in trials for not doing this, but I'm unsure how to do this apart from subbing in values and going "look its still positive!"

Hi
(Don't worry, no one saw

)

For 1 mark, I don't think you need that much proof to go with it. I would derive the acceleration formula (16e^-2t) and give the statement that e^-2t > 0 for all values of t, which makes acceleration always positive.

Hope this helps

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