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March 28, 2024, 07:35:13 pm

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jasminerulez9

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Re: Mathematics Question Thread
« Reply #4485 on: July 17, 2020, 06:30:19 pm »
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Thanks!

Coolmate

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Re: Mathematics Question Thread
« Reply #4486 on: September 05, 2020, 04:16:25 pm »
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Hi Everyone! :D

Could someone please explain why a fraction comes out the front with this integral (attached), the reasoning behind it and how to do this with any other integral?

Also, when doing revision for the HSC exam for maths, is it best to just work through the NESA past papers? Or is there anything else that would benefit my studies?

Thanks in advance!
Coolmate 8)
« Last Edit: September 05, 2020, 04:17:58 pm by Coolmate »
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Justin_L

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Re: Mathematics Question Thread
« Reply #4487 on: September 05, 2020, 04:39:06 pm »
+4
Hi Everyone! :D

Could someone please explain why a fraction comes out the front with this integral (attached), the reasoning behind it and how to do this with any other integral?

Also, when doing revision for the HSC exam for maths, is it best to just work through the NESA past papers? Or is there anything else that would benefit my studies?

Thanks in advance!
Coolmate 8)

Hey Coolmate,

-1/3 in this case is a constant, which can be removed from the integral. This can be done with any integral as the constant doesn't affect the end result since it's applied at the end anyways.

Not sure if I explained properly, hopefully this makes sense!

EDIT: With doing revision, I'm wondering that myself too. Based on my trials, my teacher actually discouraged me from doing past papers and advised that I work through revision first as to build my way up to exam level questions. Not sure if this is useful to you, hopefully someone else in this thread will have some good tips.
« Last Edit: September 05, 2020, 04:46:23 pm by Justin_L »
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Justin_L

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Re: Mathematics Question Thread
« Reply #4488 on: September 05, 2020, 05:40:59 pm »
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Heyo, would appreciate help with this question:

Simplify sinθcosθ + cosłθ/sinθ into a single trigonometric ratio

I've gotten to sinθcosθ + cotθcos˛θ, but I'm not quite sure how to go further. Any help would be appreciated, as well as any tips on solving these types of problems!
« Last Edit: September 05, 2020, 05:46:38 pm by Justin_L »
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FlammaZ

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Re: Mathematics Question Thread
« Reply #4489 on: September 05, 2020, 05:58:37 pm »
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The amount of a certain chemical in a type A cell is normally distributed with a mean of 10 and a standard deviation of 1. The amount in a type B cell is normally distributed with a mean of 14 and a standard deviation of 2. To determine whether a cell is type A or type B, the amount of chemical in the cell is measured. The cell is classified as type A if the amount is less than a specified value c, and as type B otherwise.
a-If c=12, calculate the probability that a type A cell will be misclassified, and the probability that a type B cell will be misclassified.
b-Find the value of c for which the two probabilities of misclassification are equal.
 

jamonwindeyer

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Re: Mathematics Question Thread
« Reply #4490 on: September 05, 2020, 06:57:00 pm »
+6
Heyo, would appreciate help with this question:

Simplify sinθcosθ + cosłθ/sinθ into a single trigonometric ratio

I've gotten to sinθcosθ + cotθcos˛θ, but I'm not quite sure how to go further. Any help would be appreciated, as well as any tips on solving these types of problems!

Hey! I'll try and give you my thought process. So I actually don't necessarily think your first step was a good move - Why? Because I see sines and cosines, I like this because it lets us use that great \(\sin^2+\cos^2=1\) rule. It takes a bit, but you get a bit of a gut feel to keep cosines and sines around as much as possible.

New first step - It wants it as a single ratio, so let's get a common denominator asap. Pull a \(\frac{1}{\sin{\theta}}\) out of the first term, and hopefully you can follow my steps:



These questions take practice, and my only advice is to try and get closer to what they want as soon as you can. They want a single ratio? Get a single fraction first. And in general, keep cosines and sines around as long as you can because they generally are more flexible to work with :)

jamonwindeyer

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Re: Mathematics Question Thread
« Reply #4491 on: September 05, 2020, 07:15:30 pm »
+5
The amount of a certain chemical in a type A cell is normally distributed with a mean of 10 and a standard deviation of 1. The amount in a type B cell is normally distributed with a mean of 14 and a standard deviation of 2. To determine whether a cell is type A or type B, the amount of chemical in the cell is measured. The cell is classified as type A if the amount is less than a specified value c, and as type B otherwise.
a-If c=12, calculate the probability that a type A cell will be misclassified, and the probability that a type B cell will be misclassified.
b-Find the value of c for which the two probabilities of misclassification are equal.
 

Hey! Have you tried drawing the two distributions on a graph to visualise the problem at all? I think that would be a good start. Here's the diagram I pulled up to help myself with this:



Draw yourself the two distributions for Type A and Type B, draw a dotted line at 12, and hopefully these will guide you:
- Consider just Type A for a bit. It is mean 10, std-dev of 1. 12 is two standard deviations above the mean. This means that only about 2.2% of Type A cells will have more than 12 units of this chemical. So there is a 2.2% chance that a Type A cell is misclassified.
- For Type B, it has a larger standard deviation (there is more spread in the distribution). If you had a to scale drawing provided, you'd see that the Type B curve spreads below the line at c=12 more than the Type A one spreads above it. 12 is only standard deviation below the mean of Type B cells. Therefore, you see a 15.8% chance of being misclassified.

For the last bit, you need to find where to draw the line such that the percentage of Type A cells above the line, is equal to the percentage of Type B cells below the line. Conceptually, you are looking for the value which is equally 'far' from the centre of the two distributions, when you consider that one standard deviation is larger than the other. Do you know of a numerical measure which takes into account mean and standard deviation in this way?

Have a think and if you're stuck...

CLICK ME IF STUCK
Use z scores! The misclassification rate will be the same when the z-score is the same (except one will be positive, one will be negative, because \(c\) sits on opposite sides of the mean for each distribution).



Solve for your answer :)

Hopefully this helps!

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Re: Mathematics Question Thread
« Reply #4492 on: September 05, 2020, 09:28:49 pm »
+5
Also, when doing revision for the HSC exam for maths, is it best to just work through the NESA past papers? Or is there anything else that would benefit my studies?

EDIT: With doing revision, I'm wondering that myself too. Based on my trials, my teacher actually discouraged me from doing past papers and advised that I work through revision first as to build my way up to exam level questions. Not sure if this is useful to you, hopefully someone else in this thread will have some good tips.

At this point, it's probably best to be doing papers since that is what you'll be doing 6 weeks or so down the track. However, there are still very good reasons not to be doing them as well, especially if there are holes in your understanding (based on trial results, which are a good indicator of what you need to work on in the next month or two). In general, give yourself a break and tackle light stuff (you'll have graduated real soon - enjoy yourself!) as a bare minimum (the break part is super important!) - and depending on how confident you are, try papers/extra targeted revision. Just remember not to leave doing papers until too late - there really is no substitute for those :)
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Coolmate

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Re: Mathematics Question Thread
« Reply #4493 on: September 05, 2020, 10:02:36 pm »
+1
Hey Coolmate,

-1/3 in this case is a constant, which can be removed from the integral. This can be done with any integral as the constant doesn't affect the end result since it's applied at the end anyways.

Not sure if I explained properly, hopefully this makes sense!

EDIT: With doing revision, I'm wondering that myself too. Based on my trials, my teacher actually discouraged me from doing past papers and advised that I work through revision first as to build my way up to exam level questions. Not sure if this is useful to you, hopefully someone else in this thread will have some good tips.

Thankyou Justin_L for the clarification! :)

At this point, it's probably best to be doing papers since that is what you'll be doing 6 weeks or so down the track. However, there are still very good reasons not to be doing them as well, especially if there are holes in your understanding (based on trial results, which are a good indicator of what you need to work on in the next month or two). In general, give yourself a break and tackle light stuff (you'll have graduated real soon - enjoy yourself!) as a bare minimum (the break part is super important!) - and depending on how confident you are, try papers/extra targeted revision. Just remember not to leave doing papers until too late - there really is no substitute for those :)
Thanks fun_jirachi for the advice, I really appreciate it! ;D

Coolmate 8)
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FlammaZ

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Re: Mathematics Question Thread
« Reply #4494 on: September 23, 2020, 09:56:18 pm »
0
The rate of flow of water into a tank is given by dV/dt=10e^(-t-1)(5−t) for 0≤t≤5, where V litres is the amount of water in the tank at time t minutes. Initially the tank is empty.
Find the initial rate of flow of water into the tank.
Find the value of t for which dV/dt=0.
Find the time, to the nearest second, when the rate is 1 litre per minute.
Find the first time, to the nearest second, when dV/dt<0.1.
4-Find the amount of water in the tank when t=5.
5-Find the time, to the nearest second, when there are 10 litres of water in the tank.

the last two is all i am unclear with(4,5), because when you integrate dv/dt, you get the v (the amount of water in the tank) and t( the time at which this volume is), why do you need to find the area?  note, dv/dt is 10 times e to the power of (-t-1) times (5-t)

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Re: Mathematics Question Thread
« Reply #4495 on: September 24, 2020, 01:39:33 pm »
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Not quite sure what you mean by 'why do you need to find the area' - my interpretation of your question is that you've answered it yourself (correct me if I'm wrong, though). You need to find the area under the curve defined by the rate of change as it will give you the volume/amount of water in the tank.

Reasonably sure that integral is outside the scope of the syllabus as well (unless they gave you some other prelude that would point you in that direction). Where did you get this question from? It would really help if a) you showed us what you've done already and/or b) made your request a little clearer because it'd really help us help you  - hope this makes sense :)
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FlammaZ

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Re: Mathematics Question Thread
« Reply #4496 on: September 24, 2020, 05:53:46 pm »
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Because i am new i dont really know how to use the site yet i just click on reply, sorry about that, this might not be the right area to post that kind of question. However what i am asking is this: why is it that (refer to question 4)- you must find the area under the curve, to get the liters , i thought you just had to integrate the dv/dt, which would give you v(t), then just find v(5). 

What i am trying to say is, once you integrate dv/dt, you get v(t), v representing the amount of water in the tank, and t the time in minutes. Now, number 4 is asking when t=5, what is the amount in the tank, it seems simple to just sub 5 into the equation and get the amount. Why do you have to find the area under the curve instead ? 

« Last Edit: September 24, 2020, 06:00:57 pm by FlammaZ »

fun_jirachi

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Re: Mathematics Question Thread
« Reply #4497 on: September 24, 2020, 06:15:21 pm »
+2
It never explicitly mentions area - you're confusing me now! But I think this might resolve your issue: for all intents and purposes, integration is strictly limited to a) anti-differentiation and b) finding areas under curves. Strictly speaking, finding the area under the curve defined by the rate of change is the same as integrating the equation defined by the rate of change. I'm not sure why you're bringing up area at all, to be honest - what you're thinking you 'must' do is exactly the same as what you thought you had to do in the first place (I am most definitely confused by your confusion, but hopefully this clears it up a little).

You can find the area under the curve, but you don't have to - that is the point I'm making. Your method is entirely valid. It is also valid to find the area under the curve because the time was constrained to \(0 \leq t \leq 5\). Since we're given the initial condition that the tank was empty, it's logical that \(v(5)\) is equal to the area under the curve anyway., since an upper limit was already placed on t = 5. Well within your rights to go for whichever way you choose - as long as you have valid working and the correct answer, no one will dock you marks, even if your method is different. This applies for any other question (you said: 'Why do you have to find the area under the curve instead ? ') - you NEVER have to do x method to solve y question. You can use any valid method to get to the one correct answer. Fundamentally everything runs under the same or similar principles, or they 'coincidentally' align.

Hope this helps :)
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FlammaZ

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Re: Mathematics Question Thread
« Reply #4498 on: September 24, 2020, 07:00:00 pm »
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I really appreciate your help, here are some photos of solution. Here V(5) is different to the correct answer, but the question says the tank is initially empty, so shouldn't v(0)=0, ?  confused.  :-\

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Re: Mathematics Question Thread
« Reply #4499 on: September 24, 2020, 07:06:17 pm »
+2
I really appreciate your help, here are some photos of solution. Here V(5) is different to the correct answer, but the question says the tank is initially empty, so shouldn't v(0)=0, ?  confused.  :-\

V(0)=0 is a correct assumption - the problem is there seems to be an issue with your integration. Here's a hint - just because V(0)=0, does not mean that C=0. (also, have you written x's instead of t's? This is a bit of an issue, make sure to write for the pronumeral given in the question!)

Also, your 1s look a lot like 7s - if I were you, I'd consider trying to write 1s as just a straight line, and draw 7s with a line through the middle, like so:



I feel this is just a little more obviously different than the way you're doing 1 is.