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October 30, 2020, 02:53:27 am

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Bri MT

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Re: QCE Specialist Maths Questions Thread
« Reply #15 on: March 22, 2020, 10:02:54 am »
+4
Sorry, I have another question!
I'm not quite sure how to do conditional probability for exponential probability distributions and I'm stuck on part c) of attached question.
Hopefully, that's my last question for now.
Much appreciated!

Hey,

Recall that:
$p_{x|y} = \frac{p_{X,Y} (x,y)}{p_X (x)}$

Let's think about what this means. We know that something being greater than 5 AND greater than 10, really just means it's greater than 10. So this allows us to find $p_{X,Y} (x,y)$  (use your result from step a)

Let me know if you're still confused and I'm happy to step you through it but it would be great to see your working/thoughts

A.Rose

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Re: QCE Specialist Maths Questions Thread
« Reply #16 on: June 01, 2020, 08:38:14 pm »
+3
Hi!
I have a question from Unit 4 Topic 2 and its about forces. This chapter brings in a lot of vectors and I'm quite rusty on Vectors from Unit 3 last year so I need some help with this question.
I was trying to do vector projection for this question but I was having trouble. I would greatly appreciate the help!
The question I am stuck on is part b.
Thank you!!
« Last Edit: June 02, 2020, 07:44:26 am by A.Rose »

Bri MT

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Re: QCE Specialist Maths Questions Thread
« Reply #17 on: June 02, 2020, 08:54:26 am »
+4
Hi!
I have a question from Unit 4 Topic 2 and its about forces. This chapter brings in a lot of vectors and I'm quite rusty on Vectors from Unit 3 last year so I need some help with this question.
I was trying to do vector projection for this question but I was having trouble. I would greatly appreciate the help!
The question I am stuck on is part b.
Thank you!!

Hey!

You know how for part a you can make triangles from each of the vectors to EF by adding in a vertical line, then applying the trig formulas?
For part b they've done the same thing except the triangle is red line + blue line + line perpendicular to the blue line. The angle between red line and blue line is 40 + 15.

Let me know if this clarifies things or if you still have questions

A.Rose

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Re: QCE Specialist Maths Questions Thread
« Reply #18 on: June 02, 2020, 09:46:48 am »
+3
Oh ok, Thank you.

That took me a little while to get my head around. For some reason, I had it in my head to do vector projection as the textbook example switched between vector projection and trig methods throughout.
So for these questions, you have to almost treat the direction vector as if it were horizontal and resolve the other vectors in terms of the direction vector? Sort of? Looking at it that way made sense for 11b but is there a more accurate interpretation?
Thanks again

Bri MT

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Re: QCE Specialist Maths Questions Thread
« Reply #19 on: June 02, 2020, 10:19:51 am »
+3
Oh ok, Thank you.

That took me a little while to get my head around. For some reason, I had it in my head to do vector projection as the textbook example switched between vector projection and trig methods throughout.
So for these questions, you have to almost treat the direction vector as if it were horizontal and resolve the other vectors in terms of the direction vector? Sort of? Looking at it that way made sense for 11b but is there a more accurate interpretation?
Thanks again

No worries!

You could also view the same maths I described as the dot product of the red line with a unit vector in the direction of the blue line. Both interpretations are numerically equivalent.

I think the trick here for you might be recognising that we don't want the magnitude of the red line influencing our answer, we just want to be using the direction. Therefore, if we use the dot product we use the red line divided by its magnitude, or we can use the trig method instead.

Hope this helps

Edit:

with the horizontal stuff, there's nothing in the question or any reference frame in the diagram to indicate that horizontal is meaningful or special - we could make anything "horizontal" if we wanted
« Last Edit: June 02, 2020, 10:23:55 am by Bri MT »

orla007

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Re: QCE Specialist Maths Questions Thread
« Reply #20 on: July 14, 2020, 06:16:23 pm »
+2
Hi! I just need some help with this question. Thanks in advance

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Re: QCE Specialist Maths Questions Thread
« Reply #21 on: July 14, 2020, 07:36:41 pm »
+6
Hi! I just need some help with this question. Thanks in advance
Hey orla007!
You want to find the horizontal and vertical components of the force such that the net horizontal accel is 0.25g draw an FBD first and foremost, aka as in how do you draw it or what are the forces.

Bri MT

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Re: QCE Specialist Maths Questions Thread
« Reply #22 on: July 15, 2020, 12:10:59 pm »
+4
Hey!

To build on the guidance 1729 has given, there are some key things to remember:

- the normal reaction force acts perpendicular to the surface the block is on (the table is horizontal so this force is up), and the block is initially at rest because this force is cancelling out the weight force.
- You can split up the 45 degree angle into horizontal and vertical components using trigonometry (in this case pythag works but often you'll want to use SOH CAH TOA)
- net force = mass * acceleration
- for net force you add forces in the same direction together and subtract forces in the opposite direction (i.e. frictional force is subtracted)

Have you drawn a force body diagram before?

lamh21

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Re: QCE Specialist Maths Questions Thread
« Reply #23 on: October 06, 2020, 10:41:40 pm »
0
Hi All!
I was wondering if anyone was able to potentially help me out with a Mathematical Induction Question below. Thank you!

Prove by induction that n^5− n is divisible by 240 for each odd positive integer n.

keltingmeith

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Re: QCE Specialist Maths Questions Thread
« Reply #24 on: October 07, 2020, 12:23:56 pm »
+4
Hi All!
I was wondering if anyone was able to potentially help me out with a Mathematical Induction Question below. Thank you!

Prove by induction that n^5− n is divisible by 240 for each odd positive integer n.

Yeah, so the trick with these types of induction proofs is you want to show that the LHS turns into some RHS that's equal to 240*(some number). For example, 480 is a multiple of 240 because 480=240*(2). If you were to get something disgusting out the end that looks like n^5-n=240*(n^100 - n^432 + 54432323n^2 + 5 - 45n) or something else that stupid, it's still 240*(some number), and so is still divisible by 240. Since it's an odd number, you also need to make sure your inductive proof only goes through odd numbers. There are two ways to do this:

1. Prove this is true for n=2m+1, instead of for n, or:
2. Make sure your base case is an odd number (so, start with n=1), then prove it true for n=k+2 instead of n=k+1 (WHY would this work??)

Otherwise, it's still an inductive proof at its heart. I want you to try this out for yourself first, so here's an example using an example question that you can use to see my hints in action:

Prove by induction that n3-n is divisible by 24 for all odd positive integers

So, first, the base case - n=1:

$n^3-n=1^3-1=0=24\times (0)$

Which is divisible by 24. So, step 2 - assume it's true for n=k.

... Done

Okay, step 3. Let's see if this is true for n=k+2:

$n^3-n=(k+2)^3-(k+2)\\
=k^3+6k^2+12k+8-k-2\\
=k^3-k+6k^2+12k+6\\
=k^3-k+6(k^2+2k+1)\\
=k^3-k+6(k+1)^2$

Okay, so we know that k^3-k is divisible by 24, so I'm going to substitute a 24x into there - because we don't care WHAT value it is exactly, just that it IS divisible by 24. I also know that k is an odd number, so k+1 HAS to be even - so I'm just going to call that 2y. Because again, I don't care EXACTLY what the number is, I just care about what it's divisible by. So this gives me:

$n^3-n=k^3-k+6(k+1)^2\\
=24x+6(2y)^2\\
=24x+6\times 4y^2\\
=24x+24y^2\\
=24(x+y^2)$

Which is a multiple of 24, and completes the proof

---

So, some questions I often get asked:

a) how did I know to make that expansion and factorisation in the steps?

I didn't - but it was either do that or do anything. With all of these proofs, I have no idea what direction I need to move. But, if I don't move forwards, I won't get anywhere - expanding at the start is the only thing I could do, so I did it. And every time I expand something, I expect I need to factorise it later, so when I recognise something I CAN factorise - I do it. If I factorise, and it turns out that that's NOT useful, then I can always just expand it in the next step and move on.

b) how did I know to make k+1=2y?

I didn't. All I know is, the more I can reduce things to stuff they're divisible by, the easier these proofs become - so I saw I could turn k+1 into a multiple of something, and I ran from there.

---

Also, if you're interested, here's how you'd do it using the set n=2m+1 method.

Step 1: Prove this is true for the base case, m=0:

$n^3-n=(2m+1)^3-(2m+1)\\
=(2(0)+1)^3-(2(0)+1)\\
=1^3-1=0=24(0)$

Done, simple. Now, assume this is true for m=k

... Done

Now, let's test m=k+1:

$n^3-n=(2(k+1)+1)^3-(2(k+1)+1)\\
=(2k+3)^3-(2k+3)\\
=(2k+1+2)^3-(2k+1+2)$

From here, just pick another variable (say, L=2k+1), and this follows the same steps as the one above.
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lamh21

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Re: QCE Specialist Maths Questions Thread
« Reply #25 on: October 07, 2020, 07:14:03 pm »
0
Yeah, so the trick with these types of induction proofs is you want to show that the LHS turns into some RHS that's equal to 240*(some number). For example, 480 is a multiple of 240 because 480=240*(2). If you were to get something disgusting out the end that looks like n^5-n=240*(n^100 - n^432 + 54432323n^2 + 5 - 45n) or something else that stupid, it's still 240*(some number), and so is still divisible by 240. Since it's an odd number, you also need to make sure your inductive proof only goes through odd numbers. There are two ways to do this:

1. Prove this is true for n=2m+1, instead of for n, or:
2. Make sure your base case is an odd number (so, start with n=1), then prove it true for n=k+2 instead of n=k+1 (WHY would this work??)

Otherwise, it's still an inductive proof at its heart. I want you to try this out for yourself first, so here's an example using an example question that you can use to see my hints in action:

Prove by induction that n3-n is divisible by 24 for all odd positive integers

So, first, the base case - n=1:

$n^3-n=1^3-1=0=24\times (0)$

Which is divisible by 24. So, step 2 - assume it's true for n=k.

... Done

Okay, step 3. Let's see if this is true for n=k+2:

$n^3-n=(k+2)^3-(k+2)\\
=k^3+6k^2+12k+8-k-2\\
=k^3-k+6k^2+12k+6\\
=k^3-k+6(k^2+2k+1)\\
=k^3-k+6(k+1)^2$

Okay, so we know that k^3-k is divisible by 24, so I'm going to substitute a 24x into there - because we don't care WHAT value it is exactly, just that it IS divisible by 24. I also know that k is an odd number, so k+1 HAS to be even - so I'm just going to call that 2y. Because again, I don't care EXACTLY what the number is, I just care about what it's divisible by. So this gives me:

$n^3-n=k^3-k+6(k+1)^2\\
=24x+6(2y)^2\\
=24x+6\times 4y^2\\
=24x+24y^2\\
=24(x+y^2)$

Which is a multiple of 24, and completes the proof

---

So, some questions I often get asked:

a) how did I know to make that expansion and factorisation in the steps?

I didn't - but it was either do that or do anything. With all of these proofs, I have no idea what direction I need to move. But, if I don't move forwards, I won't get anywhere - expanding at the start is the only thing I could do, so I did it. And every time I expand something, I expect I need to factorise it later, so when I recognise something I CAN factorise - I do it. If I factorise, and it turns out that that's NOT useful, then I can always just expand it in the next step and move on.

b) how did I know to make k+1=2y?

I didn't. All I know is, the more I can reduce things to stuff they're divisible by, the easier these proofs become - so I saw I could turn k+1 into a multiple of something, and I ran from there.

---

Also, if you're interested, here's how you'd do it using the set n=2m+1 method.

Step 1: Prove this is true for the base case, m=0:

$n^3-n=(2m+1)^3-(2m+1)\\
=(2(0)+1)^3-(2(0)+1)\\
=1^3-1=0=24(0)$

Done, simple. Now, assume this is true for m=k

... Done

Now, let's test m=k+1:

$n^3-n=(2(k+1)+1)^3-(2(k+1)+1)\\
=(2k+3)^3-(2k+3)\\
=(2k+1+2)^3-(2k+1+2)$

From here, just pick another variable (say, L=2k+1), and this follows the same steps as the one above.
Ahh I see! Thank you so much for your help!

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Re: QCE Specialist Maths Questions Thread
« Reply #26 on: October 09, 2020, 05:22:22 pm »
0
Hi, I’ve attempted the attached question and I’m quite stuck. I have also attached my attempted solution, however I am unsure where I have gone wrong. Thanks in advance for any help
« Last Edit: October 09, 2020, 05:37:14 pm by Adfer »

Bri MT

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Re: QCE Specialist Maths Questions Thread
« Reply #27 on: October 09, 2020, 05:28:10 pm »
0
Hi, I’ve attempted the attached question and I’m quite stuck. I have also attached my attempted solution, however I am unsure where I have gone wrong. Thanks in advance for any help

Hi,

Welcome to the forums!

I can't see any attached question, could you please edit your post and try attempting to attach it again, embedding an image of the question, or writing the question out for us? if you're unsure about how to do this there are instructions here and please feel free to ask as well

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Re: QCE Specialist Maths Questions Thread
« Reply #28 on: October 09, 2020, 05:39:04 pm »
0
Hi,

Welcome to the forums!

I can't see any attached question, could you please edit your post and try attempting to attach it again, embedding an image of the question, or writing the question out for us? if you're unsure about how to do this there are instructions here and please feel free to ask as well

Haha my apologies, took a while to attach everything - but it should be there now. I've attached the final part of my working to this post as it was too large to put all on the post above.

Britnium

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Re: QCE Specialist Maths Questions Thread
« Reply #29 on: October 17, 2020, 06:13:34 pm »
0
Hello!
I'm currently doing a PSMT on Leslie Matrices; I had believed that I've gotten a good chunk of the assignment done but after hearing other information floating around the grade I think I'm a little lost.

So our task is to model population trends of the Tasmanian devil since the documentation of the Devil Facial Tumor Disease in 1996 up until 2030 to determine whether or not the species will go extinct.

I've calculated the initial female age distributions and now I just need to develop a Leslie matrix (which is 7x7) to model the population trends. Currently, they have provided us with the following birth and survival rates, all of which are for healthy devils, so the challenge I'm facing is determining these rates for disease-affected populations.

Survival rates for rates based on historical data for disease-free populations (where s0 = probability of surviving the 0-1 age interval):

s0 = 0.39
s1 = 0.82
s2 = 0.82
s3 = 0.82
s4 = 0.82
s5 = 0.27
s6 = 0

Breeding numbers (female per female devil):

m0 = 0
m1 = 0.03
m2 = 0.86
m3 = 1.55
m4 = 1.55
m5 = 1.55
m6 - 0.86

We've also been provided with relevant research, which I'm 99% sure we're to use for developing our survival rates. Please see the screenshots attached.

My initial guess was that we were to select appropriate survival rates from the range of 0.1-0.6, model the trends using these numbers and compare the obtained populations with actual statistics (e.g. 50% killed from 1996-2007) to establish validity in the model and change the rates where necessary to match up with these figures and thus 'refine' our model. However, recently there seems to be a stress on the sentence 'a large host population will experience a rapid decline followed by stabilization and eventual return to pre-disease numbers.'  Under the assumption that birth rates will stay the same, I'm completely unsure of how to obtain survival rates that would give us this stabilization point and subsequently model the return to pre-disease numbers. And would this all be done under one Leslie matrix? Or would we expect to have multiple to model different periods of time?

Any help on this would be greatly appreciated. Sorry for such the long question and apologies if my stress has gotten to you too