September 19, 2020, 07:07:14 pm

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#### jasminerulez9

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

#### Coolmate

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« Reply #4486 on: September 05, 2020, 04:16:25 pm »
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Hi Everyone!

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?

Coolmate
« Last Edit: September 05, 2020, 04:17:58 pm by Coolmate »
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#### Justin_L

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« Reply #4487 on: September 05, 2020, 04:39:06 pm »
+3
Hi Everyone!

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?

Coolmate

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 »

#### Justin_L

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

Simplify sinθcosθ + cos³θ/sinθ into a single trigonometric ratio

I've gotten to sinθcosθ + cotθcos²θ, 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 »

#### FlammaZ

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« 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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« Reply #4490 on: September 05, 2020, 06:57:00 pm »
+5
Heyo, would appreciate help with this question:

Simplify sinθcosθ + cos³θ/sinθ into a single trigonometric ratio

I've gotten to sinθcosθ + cotθcos²θ, 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:

$
\frac{\sin^2{\theta}\cos{\theta}+\cos^3{\theta}}{\sin{\theta}}
=\frac{\cos\theta\left(\sin^2{\theta}+\cos^2{\theta}\right)}{\sin{\theta}}
=\frac{\cos{\theta}}{\sin{\theta}}
=\cot{\theta}
$

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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« Reply #4491 on: September 05, 2020, 07:15:30 pm »
+4
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).

$
\frac{c-10}{1}=-\frac{c-14}{2}
$

Hopefully this helps!

#### fun_jirachi

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« Reply #4492 on: September 05, 2020, 09:28:49 pm »
+3
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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« 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!

Coolmate
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