Author Topic: VCE Methods Question Thread! (Read 6269215 times) Tweet Share

LiquidPaperz · « **Reply #6285 on:** October 22, 2014, 11:02:50 pm »

they have said 20 do biology, so they mean 20 do just bio, how do we know if when they say Pr (A) = 0.3 for example that the value 0.3 doesnt include the intersection. So it Pr (A) is 0.3 and Pr (B) is 0.5 and Pr(a intersect b) = 0.1, from left to right would go

0.2 0.1 0.4 ??

and also when do we acknowledge the values outside the circles (is this the even space or is it Pr (A union B)' ?

theres a bit in this whole question, hope someone can decode and understand what i mean.

thanks

GeniDoi · « **Reply #6286 on:** October 22, 2014, 11:13:37 pm »

Quote from: ras on October 20, 2014, 06:13:51 pm

Last year in my bound notes I dedicated a page to my most common silly mistakes. I had a checklist of things I had to check through each question if I had enough time. It was mainly things such as units, checking working, what form of answer does the question ask for. A key point, if you have the time, is not moving on to the next question until you have reread the question you have just finished. Keep on trying to minimise silly mistakes. It may seem like you're not getting anywhere but at some point soon it should all come together

Correct me if I'm wrong but VCAA don't care at all about units. In questions that ask for units, its acceptable in Methods to leave it as a unit-less figure, even if it explicitly asks for something such as "Leave your area in sq km" (Do not attempt this for Physics, you will die.)

LiquidPaperz · « **Reply #6287 on:** October 23, 2014, 12:10:10 am »

also where would we find Pr( not A union B) on the venn diagram?

and also need help with this q

GeniDoi · « **Reply #6288 on:** October 23, 2014, 08:06:46 am »

Quote from: LiquidPaperz on October 23, 2014, 12:10:10 am

also where would we find Pr( not A union B) on the venn diagram?

and also need help with this q

I did it by first using a Karnough map and then converting it to a Venn Diagram. Sorry for dodgy paint skills:

EDIT: Got the F and G the wrong way around, fixed in this image:

keltingmeith · « **Reply #6289 on:** October 23, 2014, 08:15:39 am »

Quote from: LiquidPaperz on October 22, 2014, 11:02:50 pm

they have said 20 do biology, so they mean 20 do just bio, how do we know if when they say Pr (A) = 0.3 for example that the value 0.3 doesnt include the intersection.

If they say "P(A)", they mean the probability of getting ANY A, whether there's also a B, C, D or Z is completely irrelevant - you'll still get an A 30% of the time. If they don't want to include the intersection, they'll write $\mathbb{P}(A\B)$ or (more likely) $\mathbb{P}(A\cap B^c)$ , where $B^c$ is the complement of B.

In fact, in that first scenario you mentioned, I would've interpreted it as 20 do biology, and possibly other subjects of interest.

myanacondadont · « **Reply #6290 on:** October 23, 2014, 09:13:22 am »

How do you go about simplifying this? I thought it would become 1/x^4?

$\frac{1}{\left ( \frac{1}{x^2} \right )^2}$

keltingmeith · « **Reply #6291 on:** October 23, 2014, 09:24:33 am »

Quote from: myanacondadont on October 23, 2014, 09:13:22 am

How do you go about simplifying this? I thought it would become 1/x^4?

$\frac{1}{\left ( \frac{1}{x^2} \right )^2}$

$\frac{1}{\left ( \frac{1}{x^2} \right )^2}=\frac{1}{\left (x^{-2}\right )^2}=\frac{1}{x^{-4}}=x^4$

GeniDoi · « **Reply #6292 on:** October 23, 2014, 09:25:45 am »

Quote from: myanacondadont on October 23, 2014, 09:13:22 am

How do you go about simplifying this? I thought it would become 1/x^4?

$\frac{1}{\left ( \frac{1}{x^2} \right )^2}$

paper-back · « **Reply #6293 on:** October 23, 2014, 09:30:41 am »

Quote from: myanacondadont on October 23, 2014, 09:13:22 am

How do you go about simplifying this? I thought it would become 1/x^4?

$\frac{1}{\left ( \frac{1}{x^2} \right )^2}$

$\\\frac{1}{\left ( \frac{1}{x^{2}} \right )^{2}} \\\frac{1}{\left ( \frac{1}{x^4} \right )} \\1\cdot \frac{x^4}{1} \\\therefore x^4$

By the way,
As silly as a question this may sound like, it has caused some debate amongst my friends. Is the square root of 16, -4 and 4, or just 4?

EDIT: Beaten by Eulerfan101 and Genidoi

keltingmeith · « **Reply #6294 on:** October 23, 2014, 09:34:42 am »

Quote from: paper-back on October 23, 2014, 09:30:41 am

By the way,
As silly as a question this may sound like, it has caused some debate amongst my friends. Is the square root of 16, -4 and 4, or just 4?

It really depends. I'll show you three separate examples:

$\sqrt{16}=4 \\ -\sqrt{16}=-4 \\ x^2=16 \implies x=\pm 4$

So you see, it depends on HOW the question is asked.

Mieow · « **Reply #6295 on:** October 23, 2014, 11:06:49 am »

Is it bad if I do the du/dx substituion method that we learn in 3&4 Specialist in the methods exam when antidifferentiating?

paper-back · « **Reply #6296 on:** October 23, 2014, 12:24:58 pm »

Thanks EulerFan101
So this means that just having √16 equals to just 4 right?, not 4 AND -4

Mieow · « **Reply #6297 on:** October 23, 2014, 02:23:49 pm »

in Exam 1 of 2013 question 10c,
on my calculator $\int_{0}^{5ln(4)}\frac{3x}{10ln(4)} + 2 - 2e^{\frac{-x}{5}} dx = \frac{55}{2}ln(2) - \frac{15}{2}$ but in the assessor's report it should equal $\frac{25}{4}ln(4) - \frac{15}{2}$

Can anyone tell me what I'm doing wrong? $:-\$
I know I'm not supposed to use a calculator but I got the same answer

LiquidPaperz · « **Reply #6298 on:** October 23, 2014, 04:55:35 pm »

the probability that each of the two students doing there homework is 0.9. given independancy, what is the probability if one/only one does their hw?

keltingmeith · « **Reply #6299 on:** October 23, 2014, 06:14:28 pm »

Quote from: LiquidPaperz on October 23, 2014, 04:55:35 pm

the probability that each of the two students doing there homework is 0.9. given independancy, what is the probability if one/only one does their hw?

One student doing their homework can be modelled as Bernoulli. Since they are independent, we can then model this as Binomial with n=2 and p=0.9.

So, we work:

$\mathbb{P}(X=1)=\frac{2!}{1!1!}(0.9)^1(0.1)^1 \\ \mathbb{P}(X=1)=2*0.09=0.18$

So, the probability of only one student doing their homework is 0.09.

You could also model this in other ways, this is just a nice result of independence.

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