VCE Methods Question Thread!

[Bri MT]

I don't understand c) ii from the 2013 methods exam 
I know that you find Pr(x>4) and multiply by 200
but the question says "...expected..."- doesn't this mean that we need to find E(x)? So we use E(x)=np= 200*Pr(x>4) (getting the same ans as the line above). Even though we get the same ans, the question isn't binomial though, so maybe we can't use E(x)=np?

If you want to, you can view the number of members with that take more than 4 minutes as a binomial distribution. You can then apply E(X)=np to that.

They probably didn't consciously consider it a binomial question, and I don't either, but if it helps you understand then I don't see any harm in it, just make it clear what you are doing.


E(X) of the function is what you found in the previous part of the question, [but in a sense that was n*p aswell (n=x values p=area)]


Edit: I need to learn to type faster so VanillaRice doesn't beat me to answering

[uhoh]

Thank you so much VanillaRice and miniturtle, you rock!!

I still don't understand avg value (I know it forms a rectangle) and that the area above the average value is NOT equal to the area below the average value. But how come for this question, we just let the area under the line= area of the shaded region?

[Bri MT]

Thank you so much VanillaRice and miniturtle, you rock!!

I still don't understand avg value (I know it forms a rectangle) and that the area above the average value is NOT equal to the area below the average value. But how come for this question, we just let the area under the line= area of the shaded region?

The rectangle can exist either above or below the x-axis  (the y-value of the rectangle may be positive or negative).
In order to be 0 overall/cancel out, the above and below components of the graph must have the same size.

[nic.boal]

Completing NEAP 2017 Exam 1 today and the last thing I wanted to see was to fail on a whole question...




Answers



Part b.)
The function R(t) is stated to model the rate at which water LEAKS out. As it is a positive parabola, the only stationary point will be a minimum. The question specifically asks for the maximum rate which water leaks out. Thus, the answer must be an endpoint, but, the answer given by NEAP is the function R(t)'s minimum as given by R'(t)=0

Part c.)
This question involves solving a cubic t³-2t²+t-1872=0
I found this cubic difficult to solve in a non-calculator exam and was under the impression that such solves would not be assessable. Am I missing a glaring trick or are we expected to find t=13 by trial and error or by observation see 13 is a factor of 1872?

Part D was fine.

Any help would be greatly appreciated.

[zhen]

Completing NEAP 2017 Exam 1 today and the last thing I wanted to see was to fail on a whole question...

(Image removed from quote.)
(Image removed from quote.)

Answers
(Image removed from quote.)


Part b.)
The function R(t) is stated to model the rate at which water LEAKS out. As it is a positive parabola, the only stationary point will be a minimum. The question specifically asks for the maximum rate which water leaks out. Thus, the answer must be an endpoint, but, the answer given by NEAP is the function R(t)'s minimum as given by R'(t)=0

Part c.)
This question involves solving a cubic t³-2t²+t-1872=0
I found this cubic difficult to solve in a non-calculator exam and was under the impression that such solves would not be assessable. Am I missing a glaring trick or are we expected to find t=13 by trial and error or by observation see 13 is a factor of 1872?

Part D was fine.

Any help would be greatly appreciated.

Just did the exact same question. In my opinion, for part b, the wording was dodgy and I agree with what you said. For part c, I didn’t get the factor either, but looking back on it (gave up of the exam after I saw the equation), but I think that if we broke 1872 up into its factors and tested them out it’d be kind of doable.

[Bri MT]

Hi,

Looking over VCAA 2012 E1, For Q4c

How do you know that the statement "Daniel receives telephone calls on both Monday and Tuesday" turns in this question in to a conditional probability one?

Doesn't finding Pr(sum=4) = 0.29 already include that he receives only phone calls on 2 days since our combinations are just 2 parts (ie 1,3 or 2,2 or 3,1)?

How would this be worded if they wanted us to just find Pr(sum = 4)?

You know it is conditional because you have been given additional information about what happened, that changes the possibilities of outcomes.
You know that it changes the possibilities of outcomes because some outcomes aren't a possibility anymore (he can't receive 0, or 1 phone call total), and it makes it more likely that he will receive 2, 3, 4, 5, or 6 phone calls total.

"Find the probability that over two days Daniel received 4 phone calls" wouldn't be conditional.

[nic.boal]

Just did the exact same question. In my opinion, for part b, the wording was dodgy and I agree with what you said. For part c, I didn’t get the factor either, but looking back on it (gave up of the exam after I saw the equation), but I think that if we broke 1872 up into its factors and tested them out it’d be kind of doable.

Thanks Zhen, good luck tomorrow, heres hoping for no sketchy questions like that one!

[Rieko Ioane]

Thanks miniturtle ^^^

For part b of this question https://imgur.com/a/aFZDa
I think it's something to do with my 1st/2nd line, what am I doing wrong?

[8390]

Hey guys first time poster here, i'm extremely nervous for tomorrow, and was just curious as to what people think it will be like?

[VanillaRice]

Thanks miniturtle ^^^

For part b of this question https://imgur.com/a/aFZDa
I think it's something to do with my 1st/2nd line, what am I doing wrong?

In your first line - you have written Pr(X>7)=Pr(X<-7). This is only true if the mean is zero (i.e. so 7 will be the same distance as -7 from the mean) - in this case it is not.

Hope this helps

Hey guys first time poster here, i'm extremely nervous for tomorrow, and was just curious as to what people think it will be like?

Welcome to the forums!
Are you referring to what is expected to be on the exam? No one can tell for sure. In my opinion, it's not about being able to anticipate what type of questions will be on the exam, but rather being able to apply your knowledge to (dare I say) any type of situation. You've spent almost an entire year preparing for this - have confidence that the work and effort you have put in thus far will allow you to do your best

Good luck!

[uhoh]

The rectangle can exist either above or below the x-axis  (the y-value of the rectangle may be positive or negative).
In order to be 0 overall/cancel out, the above and below components of the graph must have the same size.

Thanks miniturtle

But for Q15 of the 2013 methods exam 2, itute said to just Select the graph which gives the same area enclosed by the
graph above or below the horizontal line y = 2 .

Doesn't that mean that if the avg value of a function is y=a, you just draw this line and the area above it= area below it?

[geminii]

Hi everyone! I'm having trouble understanding why the answer to this VCAA question is A. This doesn't work when n=1,3,5,7 and so on, because that would make an angle of 2pi/3, when the two angles are -pi/4 (or 7pi/4) and -3pi/4 (or 5pi/4). If anyone could please explain why A is the answer, that would be amazing! Thanks heaps

[Syndicate]

Hi everyone! I'm having trouble understanding why the answer to this VCAA question is A. This doesn't work when n=1,3,5,7 and so on, because that would make an angle of 2pi/3, when the two angles are -pi/4 (or 7pi/4) and -3pi/4 (or 5pi/4). If anyone could please explain why A is the answer, that would be amazing! Thanks heaps

As for n = 1, x = 3pi/4

sin(2 x 3pi/4) = sin(3p/2) = -1 (it works for all integers)
 I think you may have misinterpreted the solution to be pi - npi/4??

[geminii]

As for n = 1, x = 3pi/4

sin(2 x 3pi/4) = sin(3p/2) = -1 (it works for all integers)
 I think you may have misinterpreted the solution to be pi - npi/4??

My working out is:

sin(2x) = -1
2x = sin^-1(-1)
2x = -3pi/4, -pi/4, 5pi/4, 7pi/4, etc.
x=-3pi/8, -pi/8, 5pi/8, 7pi/8, etc.
(obviously my working leads to the wrong answer,none of the answers have values with a denominator of 8 ).

With that second step I did, I found the angles where sin = opp/hyp = -1/1 or 1/-1, which are -3pi/4, -pi/4, 5pi/4, 7pi/4, and so on. I understand what you did, you found the part of the unit circle where sin is equal to -1, which is -pi/2. Is what I did wrong? How do I know whether to do what you did or what I did?

Thanks Syndicate

EDIT: Oh my God! Just ignore me, stupid mistake....why was I thinking that opp/hyp of a pi/4 triangle is -1/1 or 1/-1? It's 1/√2! Sorry!!

[Bri MT]

Thanks miniturtle

But for Q15 of the 2013 methods exam 2, itute said to just Select the graph which gives the same area enclosed by the
graph above or below the horizontal line y = 2 .

Doesn't that mean that if the avg value of a function is y=a, you just draw this line and the area above it= area below it?

Yes, that is true. 

In this case, the average value was zero so your "cancelling out line" was at y=0
If you think about the word "average" it makes sense that the area would be the same on either side of it.

If you have for instance sin(x)  from 0 to 2pi the average value is zero (as we know that this graph is symmetrical above and below the x-axis). If you then translate that graph by a (resulting in sin(x)+a), the average value translates up by a as well, and goes from being 0 to being a.

Hope this helps you understand