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March 19, 2024, 05:58:30 pm

Author Topic: VCE Specialist 3/4 Question Thread!  (Read 2160843 times)  Share 

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Mattjbr2

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Re: Specialist 3/4 Question Thread!
« Reply #8805 on: August 20, 2017, 12:16:39 pm »
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I don't understand the reasoning behind this. Shouldn't the displacement be -10? Since displacement is relative to the origin and x(0)=0 and x(5)=-10? What did they do in the calculations? Why take away -4 and divide by 5-4?
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Shadowxo

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Re: Specialist 3/4 Question Thread!
« Reply #8806 on: August 20, 2017, 12:34:12 pm »
+3
I don't understand the reasoning behind this. Shouldn't the displacement be -10? Since displacement is relative to the origin and x(0)=0 and x(5)=-10? What did they do in the calculations? Why take away -4 and divide by 5-4?
I think that's just a really bad question plus an error on their part
I think what they meant to ask was "what's the average velocity between t=4 and t=5" but didn't (which is where the 5-4 would come from)
Or they meant the displacement at t=5 relative to t=4 (but that wouldn't use the 5-4)

You're correct, it's an error in the book so don't worry about it! :)
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Willba99

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Re: Specialist 3/4 Question Thread!
« Reply #8807 on: August 20, 2017, 03:30:11 pm »
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If you want to find its integral, it's not a standard integral so I believe you'd have to do partial fractions and integrate it that way.
If you want to find the derivative, just use the chain rule

Sorry I meant anti-derive.
I tried the partial fractions and i cant seem to separate them right. I keep getting 1/2sqrt(a) as both my A and B values, but when i graph the partial fraction function compared to the original one (using arbitrary values for a and b) they aren't the same. Could you (or anyone) have a look at my working out? I attached it as a picture.
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Shadowxo

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Re: Specialist 3/4 Question Thread!
« Reply #8808 on: August 20, 2017, 03:42:16 pm »
+2
Sorry I meant anti-derive.
I tried the partial fractions and i cant seem to separate them right. I keep getting 1/2sqrt(a) as both my A and B values, but when i graph the partial fraction function compared to the original one (using arbitrary values for a and b) they aren't the same. Could you (or anyone) have a look at my working out? I attached it as a picture.
Your working out is fine - I graphed it to make sure and they're appearing as the same graph :) maybe double checked you typed it in right
Nice Job :D
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Willba99

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Re: Specialist 3/4 Question Thread!
« Reply #8809 on: August 20, 2017, 03:45:32 pm »
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Your working out is fine - I graphed it to make sure and they're appearing as the same graph :) maybe double checked you typed it in right
Nice Job :D

Thanks very much!!
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Syndicate

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Re: Specialist 3/4 Question Thread!
« Reply #8810 on: August 20, 2017, 05:03:02 pm »
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Hey,

Can someone please help me with this:
An alarm system has 20 batteries that are connected so that, when one battery fails, the next one takes over. (Only one battery is working at any one time.) The batteries operate independently, and each has a mean life of 7 hours and a standard deviation of 0.5 hours. What is the probability that the alarm system is still working after 145 hours?

Working out:
E(20X) = 140

sd(20X) = sqrt{400 x 0.5^2) = 10 (I only get the right answer, when I multiply the variance by 20, not 400 (I thought Var(aX) = a^2Var(X)? ))

Z = (145-140)/10 = 0.5
Pr(Z>0.5) = 0.30854

(The right answer is 0.01267)

Just to clarify: I am asking why do we have to multiply the Variance by 20, why not 400?

« Last Edit: August 20, 2017, 05:10:40 pm by Syndicate »
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VanillaRice

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Re: Specialist 3/4 Question Thread!
« Reply #8811 on: August 20, 2017, 05:17:04 pm »
+2
Hey,

Can someone please help me with this:
An alarm system has 20 batteries that are connected so that, when one battery fails, the next one takes over. (Only one battery is working at any one time.) The batteries operate independently, and each has a mean life of 7 hours and a standard deviation of 0.5 hours. What is the probability that the alarm system is still working after 145 hours?

IMPORTANT EDIT 2: Please see my comment (2 comments below) for a correction and clarification.

We'll let X = the length of time one battery can run
And Y = X1 + X2 ... X20 = 20X
E(Y) = 7 x  20 = 140

Var(X) = 0.52 = 0.25
Var(Y) = 0.25 x 20 = 5
SD(Y) = 2.236 (square root of variance)

Pr(Y > 145) = ?

A problem you may have run into is converting SD(X) into SD(Y) - since we don't have a rule for the standard deviation of linear combinations, we must first convert SD into variance.
Hope this helps :)

EDIT: [REMOVED - what I said was confusing and irrelevant]
« Last Edit: August 20, 2017, 05:51:54 pm by VanillaRice »
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Syndicate

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Re: Specialist 3/4 Question Thread!
« Reply #8812 on: August 20, 2017, 05:30:52 pm »
+1
We'll let X = the length of time one battery can run
And Y = X1 + X2 ... X20 = 20X
E(Y) = 7 x  20 = 140

Var(X) = 0.52 = 0.25
Var(Y) = 0.25 x 20 = 5
SD(Y) = 2.236 (square root of variance)

Pr(Y > 145) = ?

A problem you may have run into is converting SD(X) into SD(Y) - since we don't have a rule for the standard deviation of linear combinations, we must first convert SD into variance.
Hope this helps :)

EDIT: Just saw your edit - recall the square of SD is variance. Since we're dealing with SD (not variance) here, we multiply by 202 = 400 :) In my opinion, it would be safer to convert to variance.

Hi,

Thanks for your reply, but I am still unsure what you really mean by your last few sentences. 

shouldn't Var(Y) = Var(20X) = 400Var(X)?

Mikki buys three bags of bananas and two bags of apples from the greengrocer. If bags of bananas have a mean weight of 750 g, with a variance of 25, and bags of apples have a mean weight of 1000 g, with a variance of 50, what are the mean and standard deviation of the total weight of her ]purchases?
That formula worked out perfectly fine for this question^.
« Last Edit: August 20, 2017, 05:33:05 pm by Syndicate »
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VanillaRice

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Re: Specialist 3/4 Question Thread!
« Reply #8813 on: August 20, 2017, 05:41:41 pm »
+3
Hi,

Thanks for your reply, but I am still unsure by your last few sentence.

shouldn't Var(Y) = Var(20X) = 400Var(X)?
Whoops! I think I confused myself there  :-X
Apologies, please ignore that last sentence (actually, you can probably ignore that entire comment  :P) I've written something that may have confused you (and myself).

Let's start again:
The question states that the batteries are independent of each other.
So, Y =  X1 + X2 + ... + X20
Note, Y = 20X is NOT true, each 'X' is independent of the next (this is where I went wrong last time).

Recall that for independent variables A and B,
Var(A + B) = Var(A) + Var(B)
i.e. we add the individual variances together

Therefore,
Var(Y) = Var(X1 + X2 + ... + X20) = Var(X1) + Var(X2) + .... + Var(X20).
We are adding the individual variances together.

Note the difference between the above, and Var(Y) = Var(20X). In the above, we are adding each individual variances together, which is why we multiply the variance by 20.

I hope this clears things up for you, and apologies for any confusion  :)



« Last Edit: August 20, 2017, 06:27:24 pm by VanillaRice »
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Syndicate

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Re: Specialist 3/4 Question Thread!
« Reply #8814 on: August 20, 2017, 05:55:59 pm »
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Whoops! I think I confused myself there  :-X
Apologies, please ignore that last sentence (actually, you can probably ignore that entire comment  :P) I've written something that may have confused you (and myself).

Let's start again:
The question states that the batteries are independent of each other.
So, Y =  X1 + X2 ... X20
Note, Y = 20X is NOT true, each 'X' is independent of the next (this is where I went wrong last time).

Recall that for independent variables A and B,
Var(A + B) = Var(A) + Var(B)
i.e. we add the individual variances together

Therefore,
Var(Y) = Var(X1 + X2 ... X20) = Var(X1) + Var(X2) + .... + Var(X20).
We are adding the individual variances together.

Note the difference between the above, and Var(Y) = Var(20X). In the above, we are added each individual variances together, which is why we multiply the variance by 20.

I hope this clears things up for you, and apologies for any confusion  :)





haha thanks so much, this makes a lot more sense now :D
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avocadoxxxx

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Re: Specialist 3/4 Question Thread!
« Reply #8815 on: August 20, 2017, 08:00:18 pm »
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For the first attachment, I get that you multiply the two vectors and equate the product to 0, but I don't understand why you reject t=1 (the answer is t=0 or t=2)

For the second attachment, why is one cycle 1 unit of time and not 2 units (4pi/2pi)? If, for example, it had been cos(3pi t) instead of cos(2pi t), what would be the period/cycle then?

Thanks!

Shadowxo

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Re: Specialist 3/4 Question Thread!
« Reply #8816 on: August 20, 2017, 08:22:43 pm »
+3
For the first attachment, I get that you multiply the two vectors and equate the product to 0, but I don't understand why you reject t=1 (the answer is t=0 or t=2)

For the second attachment, why is one cycle 1 unit of time and not 2 units (4pi/2pi)? If, for example, it had been cos(3pi t) instead of cos(2pi t), what would be the period/cycle then?

Thanks!
First picture: The reason they didn't have t=1 as a solution is because that gives a zero vector (0i + 0j) so therefore cannot be "perpendicular" to anything.
Second: the period is 2pi/n, so for the first term, cos(2πt), the period is 2π/2π = 1. For the second term, cos(4πt), period is 2π/4π = 1/2
Hope this clears things up :)
« Last Edit: August 20, 2017, 10:10:15 pm by Shadowxo »
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avocadoxxxx

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Re: Specialist 3/4 Question Thread!
« Reply #8817 on: August 20, 2017, 08:29:52 pm »
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First picture: The reason they didn't have t=1 as a solution is because that gives a zero vector (0i + 0j) so therefore cannot be "perpendicular" to anything.
Second: the period is n/2pi, so for the first term, cos(2πt), the period is 2π/2π = 1. For the second term, cos(4πt), period is 4π/2π = 2
Hope this clears things up :)

Thanks for that! But for the second question, the answer says that the particle (its position is given by the vector) completes one cycle in one unit of time (not 2 units) - I was wondering why this is the case? How do you know which term to go by? (Sorry, probably should've been clearer with my question) 

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Re: Specialist 3/4 Question Thread!
« Reply #8818 on: August 20, 2017, 08:42:47 pm »
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Hi,

I need some help on these 2 MCQ Qs https://m.imgur.com/a/SOIYx

Q5) isn't A also correct in addition to C? From methods, can't we approximate binomial distributions as normal so A?

Q7) Couldn't D also be correct in addition to B?

VanillaRice

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Re: Specialist 3/4 Question Thread!
« Reply #8819 on: August 20, 2017, 09:01:41 pm »
+3
Hi,

I need some help on these 2 MCQ Qs https://m.imgur.com/a/SOIYx

Q5) isn't A also correct in addition to C? From methods, can't we approximate binomial distributions as normal so A?

Q7) Couldn't D also be correct in addition to B?
5) Recall the definition of the central limit theorem: given a large sample size, the distribution of sample means of a population is normal
Option A implies that ALL distribution are normal, provided the sample size is large enough. While this may be true in your case of binomial distributions, it does not always hold. Take, for example, the income distribution in a region of low socioeconomic class. The data would have most of the population at low income, while a few people will have high incomes. This is not normally distributed.

7) Option D is not the definition of a sample distribution of sample means. The sampling distribution of sample means is the distribution of the means of many samples taken from a population.

It might take awhile to get your head around all these terms which sound the same, but hopefully this helps :)

EDIT: I find that this video is helpful is explaining the central limit theorem, and also why the sampling distribution of sample means is normal, regardless of the distribution of the population.
« Last Edit: August 20, 2017, 09:08:46 pm by VanillaRice »
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