 November 13, 2019, 01:09:30 am

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#### not a mystery mark « Reply #4395 on: October 22, 2019, 11:19:09 am »
0
Not a calculation question, but a HSC maths marking question.
If you get a 3 mark question correct but provide no working - will you get the 3 marks or will points be deducted?

Thanks heaps to the og who answers this <3 #### RuiAce

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• Respect: +2390 « Reply #4396 on: October 22, 2019, 11:26:14 am »
+1
Not a calculation question, but a HSC maths marking question.
If you get a 3 mark question correct but provide no working - will you get the 3 marks or will points be deducted?

Thanks heaps to the og who answers this <3
For a 3 marker, working out is definitely mandatory.

If you get the final answer correct with no working, the most likely scenario is that you'd get 1/3. (2/3 and 0/3 would be in exceptional circumstances in my opinion.)  #### not a mystery mark « Reply #4397 on: October 22, 2019, 01:35:35 pm »
0
For a 3 marker, working out is definitely mandatory.

If you get the final answer correct with no working, the most likely scenario is that you'd get 1/3. (2/3 and 0/3 would be in exceptional circumstances in my opinion.)

Legendary!! Thank you heaps. You have legendary OG status. #### s.jay

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• Respect: 0 « Reply #4398 on: October 22, 2019, 09:19:49 pm »
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Hey everyone,
I know this may sound silly but with days before my maths exams, and after prepping extensively on the more difficult topics, I realised today that I forgot how to do questions with bearings (as in prelim trig). I know that I need to use sine and cosine rule etc. but I have forgotten how to actually find the angles!
If anyone could give me a quick refresher that would be great!!

Also here is a question that I am stuck with:
Sally starts on J and swims on a bearing of 105 deg for 5km out to L. She then changes direction and swims on a bearing of 045 deg for a further 16km.
(i) find angle JLM
(ii) find the distance JM that Sally will swim back to J

Thanks so much #### fun_jirachi

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• Respect: +269 « Reply #4399 on: October 22, 2019, 10:51:01 pm »
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Hey there! Welcome to the forums!

Always a good idea to draw a diagram; that should always be your first step. I've put it in the spoiler below so this post takes up less space, but please take a look at it, it will go a long way to helping you understand any bearing questions that come your way Spoiler I've actually forgotten to put in the 45 degree angle, but from the information given in the question and angle sum of a triangle, you should be able to deduce that angle JLM = 120 degrees.
And for part b), it's just one application of the cosine rule, with sides 5 and 16, with enclosed angle 120 degrees as found in part a). Have a go at more questions - starting with drawing a diagram! While you might be able to visualise it well, exams aren't really the places to hedge bets, it's better to draw them out always Hope this helps Failing everything, but I'm still Flareon up.

HSC 2018: Modern History  | 2U Maths 
HSC 2019: Physics | Chemistry | English Advanced | Maths Extension 1 | Maths Extension 2

#### Kombmail « Reply #4400 on: October 22, 2019, 11:08:18 pm »
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Guys how do you find solutions to tan x =-2?
-KgkG-

#### Hawraa « Reply #4401 on: October 23, 2019, 01:35:50 pm »
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Guys how do you find solutions to tan x =-2?

Hi,
I'd do it this way. Can u confirm if the answer is correct ☺️?

#### Hawraa « Reply #4402 on: October 23, 2019, 01:46:06 pm »
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This is a multiple choice question from 2015 : which expression is a term of the geometric series 3x-6x^2+12x^3-...?
A)3072x^10
B)-3072x^10
C)3072x^11
D)-3072x^11

But how would you do this question?

Hi there,

#### RuiAce

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• Respect: +2390 « Reply #4403 on: October 23, 2019, 01:47:52 pm »
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Hi,
I'd do it this way. Can u confirm if the answer is correct ☺️?
Their question was unclear because they didn't specify the domain for $\theta$.

If we assume that $0^\circ \leq \theta \leq 360^\circ$, then your method is certainly correct. (Of course, in the exam, they could ask for radians instead.)  #### mani.s_

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• Respect: 0 « Reply #4404 on: October 23, 2019, 05:14:11 pm »
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Hi, I just wanted to know the difference between standard deviation and variance? I know that standard deviation measures the spread of data from the mean, so 68% of the data will be in 1 standard deviation and 95% of the data will be in 2 standard deviations. I was also told that variance is the same thing as standard deviation except its squared. So what's the point of having variance and how is it actually different to standard deviation and how is it useful in maths?

Thanks for your time to answer my question #### RuiAce

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• Respect: +2390 « Reply #4405 on: October 23, 2019, 05:47:35 pm »
+1
Hi, I just wanted to know the difference between standard deviation and variance? I know that standard deviation measures the spread of data from the mean, so 68% of the data will be in 1 standard deviation and 95% of the data will be in 2 standard deviations. I was also told that variance is the same thing as standard deviation except its squared. So what's the point of having variance and how is it actually different to standard deviation and how is it useful in maths?

Thanks for your time to answer my question The standard deviation is nothing more but the square root of the variance.
$SD(X) = \sqrt{\operatorname{Var}(X)}$
Why do we have both? Because in the context of random variables, the absence of the square root makes the formula look nicer.
$\operatorname{Var}(X) = E\left[(x-\mu)^2\right]$
That, and it's also more cleaner to use in more advanced (university level) mathematical statistics proofs.

But once the results are derived, it's usually more of interest to report standard deviations to the public. The variance rescales things according to square of quantities, whilst the standard deviation is on the same scale.

Edit: Upon looking at the question again, I guess the main answer is that the variance is more commonly used for proofs in mathematical statistics at university. The variance was the first convention used by mathematicians, and a lot of the theory of statistics has been developed based off it. For many famous distributions, the variance generally takes a nicer form - the standard deviation introduces a square root out of nowhere. Sure, a square root isn't overly harmful or anything, but it's just nicer to not have it there altogether. Also, computers may run into precision errors when dealing with square roots, as opposed to just positive integer powers.

But for summary statistics, I'd probably say one would be crazy to only use the variance instead of the SD.
« Last Edit: October 23, 2019, 06:03:19 pm by RuiAce »  #### mani.s_

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• Respect: 0 « Reply #4406 on: October 23, 2019, 06:04:28 pm »
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The standard deviation is nothing more but the square root of the variance.
$SD(X) = \sqrt{\operatorname{Var}(X)}$
Why do we have both? Because in the context of random variables, the absence of the square root makes the formula look nicer.
$\operatorname{Var}(X) = E\left[(x-\mu)^2\right]$
That, and it's also more cleaner to use in more advanced (university level) mathematical statistics proofs.

But once the results are derived, it's usually more of interest to report standard deviations to the public. The variance rescales things according to square of quantities, whilst the standard deviation is on the same scale.
So pretty much the variance and standard deviation are the exact same thing, except that variance is used to make the formulas look nicer and
working out more cleaner.

#### mani.s_

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• Respect: 0 « Reply #4407 on: October 23, 2019, 06:08:59 pm »
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Rui one more thing, for the Expected value of binomial distribution E(X) = np and Variance Var(X) = np(1-p), do we learn how to derive these formulas in school, cause I havent really seen anything in the books and if not, can u please show me how to derive these formulas???

Thanks #### RuiAce

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• Respect: +2390 « Reply #4408 on: October 23, 2019, 06:30:29 pm »
+1
So pretty much the variance and standard deviation are the exact same thing, except that variance is used to make the formulas look nicer and
working out more cleaner.
Yep.
Rui one more thing, for the Expected value of binomial distribution E(X) = np and Variance Var(X) = np(1-p), do we learn how to derive these formulas in school, cause I havent really seen anything in the books and if not, can u please show me how to derive these formulas???

Thanks These proofs are actually heavily involved. I had trouble doing them in first year; wasn't until second year university when I started finding it easy. Here's the proof for the expected value at least just for viewing pleasure.

(Note: So therefore you definitely won't need it at the high school level.)
\begin{align*}
E(X) &= \sum_x x P(X=x)\\
&= \sum_{x=0}^n x \binom{n}{x}p^x (1-p)^{n-x}\\
&= \sum_{x=0}^n x \frac{n!}{x!(n-x)!}p^x (1-p)^{n-x}.
\end{align*}
Now, we first evaluate the sum at $x=0$. We pull the value when $x=0$ outside of the sum.
\begin{align*}
E(X) &= 0 \binom{n}{0}p^0 (1-p)^{n-0} + \sum_{x=1}^n x\binom{n}{x}p^x (1-p)^{n-x}\\
&= \sum_{x=1}^n x\binom{n}{x}p^x (1-p)^{n-x}
\end{align*}
because that first term actually evaluates to 0. Now we note that the terms in the sum are indexed by $x=1, 2, 3, \dots, n$. So every term in the sum now always has the property $x=0$, and hence we can cancel any $x$'s in the numerator and denominator.

In addition to this, we need the trick $N! = N(N-1)!$. (You can try to convince yourself that this formula is true by simply expanding both of the factorials.) Here, I first use that $x! = x(x-1)!$. (Note: The assumption for this question is that $x$ is an integer; not any real number.)
\begin{align*}
E(X)& = \sum_{x=1}^n x\binom{n}{x}p^x (1-p)^{n-x}\\
&= \sum_{x=1}^n x \frac{n!}{x!(n-x)!} p^x (1-p)^{n-x}\\
&= \sum_{x=1}^n \frac{n!}{(x-1)(n-x)!} p^x (1-p)^{n-x}
\end{align*}
Observe that we had to tear apart the binomial coefficient (i.e. convert it to its factorial notation) to cancel out the $x$ in front. We now have to rebuild a new binomial coefficient using the remaining factorials. It turns out that we now need to manipulate it to obtain $\binom{n-1}{x-1}$.

This requires that we use $n! = n(n-1)!$ in the numerator. Once we do that, observe that the sum is in terms of $x$, and not in terms of $n$. Hence the extra $n$ can now be moved in front.
\begin{align*}
E(X) &= \sum_{x=1}^n \frac{n(n-1)!}{(x-1)!(n-x)!}p^x (1-p)^{n-x}\\
&= n \sum_{x=1}^n \frac{(n-1)!}{(x-1)!(n-x)!} p^x (1-p)^{n-x}\\
&= n \sum_{x=1}^n \frac{(n-1)!}{(x-1)![(n-1)-(x-1)!]} p^x (1-p)^{n-x}\\
&=  n \sum_{x=1}^n \binom{n-1}{x-1}p^x (1-p)^{n-x}
\end{align*}
Now I'll decompose $p^x$ into $p\times p^{x-1}$, which allows me to pull out a factor of $p$ in front as well.
\begin{align*}
E(X) &=  n \sum_{x=1}^n\binom{n-1}{x-1} p\, p^{x-1}(1-p)^{n-x}\\
&= np \sum_{x=1}^n \binom{n-1}{x-1} p^{x-1}(1-p)^{n-x}
\end{align*}
__________________________________________________________________

Observe that our end goal $np$ has now showed up. It remains to prove that the ugly sum actually equals to 1. To do this, I will now tear apart the sum, by subbing every value of $x$ from $1$ to $n$ in. I'll focus only on the sum here.
\begin{align*}
&= \binom{n-1}{1-1} p^{1-1} (1-p)^{n-1} + \binom{n-1}{2-1} p^{2-1}(1-p)^{n-2} + \binom{n-1}{3-1} p^{3-1}(1-p)^{n-3} + \cdots + \binom{n-1}{n-1} p^{n-1} (1-p)^{n-n}\\
&= \binom{n-1}{0} (1-p)^{n-1} + \binom{n-1}{1} p(1-p)^{(n-1)-1} + \binom{n-1}{2} p^2 (1-p)^{(n-1)-2} + \cdots + \binom{n-1}{n-1} p^{n-1}
\end{align*}
It may or may not be obvious here, but this now looks disturbing like the statement of the binomial theorem!
Recall: Statement of the binomial theorem
$(x+y)^n = \binom{n}{0} x^n + \binom{n}{1} x^{n-1}y + \binom{n}{2} x^{n-2} y^2 + \cdots + \binom{n}{n}y^n.$
Here, the power should be to $n-1$ instead. Observe how $p$ replaces the role of $x$, and $1-p$ replaces the role of $y$.

So from using the binomial theorem, that expression now simplifies to
$\left[ p + (1-p)\right]^{n-1}.$
Hopefully it is clear that this expression indeed evaluates to 1. Thus upon substituting into what we had earlier,
$E(X) = np \times 1 = np$
as required.

For the variance, one uses a similar strategy to this, but they start by computing $E(X(X-1))$ instead. Since $E(X(X-1)) = E(X^2 - X) = E(X^2) - E(X)$, this then allows us to find the second moment $E(X^2)$. And as usual, conclude with $\operatorname{Var}(X) = E(X^2) - [E(X)]^2$.
« Last Edit: October 23, 2019, 06:33:56 pm by RuiAce »   