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March 29, 2024, 09:22:07 am

Author Topic: 3U Maths Question Thread  (Read 1230405 times)  Share 

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RuiAce

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Re: 3U Maths Question Thread
« Reply #4020 on: April 08, 2019, 10:00:53 am »
+3
Looking at it now I reckon it needs some backtracking.
\[ \text{We expect that on any random day,}\\ \text{the probability of at least 1 machine not working is }0.8.\\ \text{Hence we should expect that on any random day,}\\ \text{the probability that all machines work is }0.2. \]
\[ \text{Thus if }p\text{ is the probability that a machine works,}\\ \binom{16}{16} p^{16} = 0.2 \implies \boxed{p = 0.2^{1/16}}.\]
So we expect that the probability of a randomly selected computer to not work is just \( 1 - 0.2^{1/16}\). You should also be able to do part b) now.

terassy

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Re: 3U Maths Question Thread
« Reply #4021 on: April 08, 2019, 08:03:39 pm »
0
Looking at it now I reckon it needs some backtracking.
\[ \text{We expect that on any random day,}\\ \text{the probability of at least 1 machine not working is }0.8.\\ \text{Hence we should expect that on any random day,}\\ \text{the probability that all machines work is }0.2. \]
\[ \text{Thus if }p\text{ is the probability that a machine works,}\\ \binom{16}{16} p^{16} = 0.2 \implies \boxed{p = 0.2^{1/16}}.\]
So we expect that the probability of a randomly selected computer to not work is just \( 1 - 0.2^{1/16}\). You should also be able to do part b) now.

Thanks again :) For the second part this is what I did:

Is this correct?

iktimal

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Re: 3U Maths Question Thread
« Reply #4022 on: April 08, 2019, 08:37:02 pm »
0
Hey guys,
Can someone help me with this probability question:
1.
a) In how many ways can a committee of 4 men and 3 women be chosen from 7 men and 8 women if two particular women refuse too serve together?
b) What is the probability that the 2 women do not serve together?

Thanks.

fun_jirachi

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Re: 3U Maths Question Thread
« Reply #4023 on: April 08, 2019, 09:00:56 pm »
+1
Hey there!

For a), there are two different ways you could do this;
i) considering the complement (total number of ways - both women in)
ii) going case by case

Method 1: Total number of ways = 7C4 x 8C3 (hopefully self explanatory)
Both women in: 2C2 (to get both women in) x 6C1 (for the other woman) x 7C4 (for the men)
= 1750.
Method 2: The cases are going to be when one woman serves, or neither woman serves.
When one woman serves: 2C1 (pick one of the women) x 6C2 (for the rest of the women, omitting the other one) x 7C4 (for the men)
When neither woman serves: 6C3 (for the women, omitting the two) x 7C4 (for the men)
Add them up, you should have 1750.

For part b), just use your answer from a), and divide by the total number of ways (outlined in method 1.) Answer should be 25/28 :)

Hope this helps :)
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philip_depasquale

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Re: 3U Maths Question Thread
« Reply #4024 on: April 09, 2019, 05:57:29 pm »
0
Hey guys
Can someone please help me with this modified natural growth and decay question. It's question 10 of exercise 7H from the Cambridge Extension I Math Textbook.
Any help would be greatly appreciated  :)

RuiAce

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Re: 3U Maths Question Thread
« Reply #4025 on: April 09, 2019, 06:12:20 pm »
+1
Hey guys
Can someone please help me with this modified natural growth and decay question. It's question 10 of exercise 7H from the Cambridge Extension I Math Textbook.
Any help would be greatly appreciated  :)
Considering this question has so many parts, can you provide more details on where you need help with, or if possible which parts specifically? Or alternatively provide any progress you've made thus far?
Thanks again :) For the second part this is what I did:
(Image removed from quote.)
Is this correct?
Looks right

_Himani_

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Re: 3U Maths Question Thread
« Reply #4026 on: April 10, 2019, 06:32:48 pm »
0
Hello,
Can someone please help me with this question: show v^2 = 48-3x^2 is a simple harmonic motion by using differentiation.
I'm not sure how to start it off; should I make x the subject and do dv/dx?
Thanks, and hope everyone's having a great week!

RuiAce

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Re: 3U Maths Question Thread
« Reply #4027 on: April 10, 2019, 06:51:24 pm »
+1
Hello,
Can someone please help me with this question: show v^2 = 48-3x^2 is a simple harmonic motion by using differentiation.
I'm not sure how to start it off; should I make x the subject and do dv/dx?
Thanks, and hope everyone's having a great week!

Hint: You're supposed to use the formula \( \frac{d^2x}{dt^2} = \frac{d}{dx} \left( \frac12 v^2\right)\). (Note that the formula differentiates with respect to \(x\), i.e. the displacement, and not with respect to the time \(t\).)

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Re: 3U Maths Question Thread
« Reply #4028 on: April 10, 2019, 07:39:07 pm »
0
(Apologies for posting in the wrong forum, I am a 2U maths student but I think my teacher may have thrown a 3U question into my homework)

The graph given in this question is y = f(x) and it wants you to sketch y = f^-1(x)

RuiAce

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Re: 3U Maths Question Thread
« Reply #4029 on: April 10, 2019, 07:45:56 pm »
+1
(Apologies for posting in the wrong forum, I am a 2U maths student but I think my teacher may have thrown a 3U question into my homework)

The graph given in this question is y = f(x) and it wants you to sketch y = f^-1(x)
Yeah, it certainly should not be placed among 2U material. Inverse functions is a MX1 topic in its entirety.
\[ \text{Restricting our focus to }x\geq 0\\ \text{we note that the single stationary point in this domain is at }x=0. \]
\[ \text{Hence the largest possible domain restriction that will work}\\ \text{is literally }x \geq 0\text{ itself}.\]
i.e. We should consider everything to the right of the \(y\)-axis.
\[ \text{From here, the graph is just obtained}\\ \text{by flipping everything in the domain }x\geq 0\\ \text{about the line }\boxed{y=x}\text{, which has been provided.}\]
Note that since (1,1) lies on \(y=f(x)\), it must lie on \(y=f^{-1}(x)\) as well.

avocadinq

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Re: 3U Maths Question Thread
« Reply #4030 on: April 12, 2019, 12:32:39 pm »
0
Hey there, just wondering how would I do q9. (f)? Thanks in advance.
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jamonwindeyer

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Re: 3U Maths Question Thread
« Reply #4031 on: April 12, 2019, 03:20:01 pm »
+1
Hey there, just wondering how would I do q9. (f)? Thanks in advance.

Hey! So the derivative of the logarithm is:



So now we just need that top derivative, which is just the chain rule!!



Put all that together, it should simplify a tad ;D

RuiAce

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Re: 3U Maths Question Thread
« Reply #4032 on: April 12, 2019, 05:21:33 pm »
+2
Hey! So the derivative of the logarithm is:



So now we just need that top derivative, which is just the chain rule!!



Put all that together, it should simplify a tad ;D
Personal opinion but I feel like log laws are nicer here though, simplifies most of the stuff for you
\begin{align*}\frac{d}{dx} \ln \sqrt{\sin^{-1}x} &= \frac12 \frac{d}{dx} \ln \left( \sin^{-1}x\right) \\ &= \frac12 \frac{\frac{1}{\sqrt{1-x^2}}}{\sin^{-1}x} \end{align*}

Jefferson

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Re: 3U Maths Question Thread
« Reply #4033 on: April 13, 2019, 10:58:06 pm »
0
HI all, could someone please explain this question to me, since I don't quite see where the 'method fails' in this scenario (attachment 1).
My calculation for the first approximation landed me closer to the root, and all subsequent approximations (attachment 2) also seem to approach the root.


RuiAce

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Re: 3U Maths Question Thread
« Reply #4034 on: April 14, 2019, 09:21:41 am »
0
HI all, could someone please explain this question to me, since I don't quite see where the 'method fails' in this scenario (attachment 1).
My calculation for the first approximation landed me closer to the root, and all subsequent approximations (attachment 2) also seem to approach the root.


Seems fine to me too.

Where did you get this question? It looks faulty.