October 14, 2019, 05:05:19 pm

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hums_student

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« Reply #18135 on: September 18, 2019, 11:43:56 am »
+1
Essentially, what I am saying is, is it POSSIBLE to get a 25+ study score if you fail a single SAC?

It depends on a lot of other factors but to put simply, it's definitely possible. I failed a methods SAC last year and still got a 34, and I was in a below-average cohort
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milanander

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« Reply #18136 on: September 18, 2019, 08:47:37 pm »
0
Just out of sheer curiosity, what would happen if your CAS dies early on during the exam?
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« Reply #18137 on: September 21, 2019, 11:01:38 am »
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Not sure how to approach the following question
I understand how to do a and b but is there an easier way to do c? It seems way to tedious to do by hand.

A company employs a sales team of 20 people, consisting of 12 men and 8 women
a) what is p, the proportion of men in the sales team
b) Five salespeople are to be selected at random to attend an important conference. What are the possible values of the same proportion ^p of men in the sample.
c) Construct a probability distribution table which summarises the sampling distribution of the sample proportion of men when sample of size 5 are selected from the sales team.

(Chapter 17B q2 from Cambridge Mathematical Methods Units 3&4 book )

aspiringantelope

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« Reply #18138 on: September 21, 2019, 03:22:53 pm »
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HARD QUESTION !!!
Two signposts are 100 km apart on the tollway.
There are six complete sections of road between these two signposts.
The lengths of the successive sections of road increase by 5%.
Determine the length of the first section of road.

THANKS
Irrelevant but according to your signature, haven't you completed VCE?
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TrueTears

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« Reply #18139 on: September 23, 2019, 06:12:54 am »
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Not sure how to approach the following question
I understand how to do a and b but is there an easier way to do c? It seems way to tedious to do by hand.

A company employs a sales team of 20 people, consisting of 12 men and 8 women
a) what is p, the proportion of men in the sales team
b) Five salespeople are to be selected at random to attend an important conference. What are the possible values of the same proportion ^p of men in the sample.
c) Construct a probability distribution table which summarises the sampling distribution of the sample proportion of men when sample of size 5 are selected from the sales team.

(Chapter 17B q2 from Cambridge Mathematical Methods Units 3&4 book )
Here, you can calculate the exact sampling distribution by hand (yes it's tedious). If the sample size was much larger, then the sampling distribution can be approximated by a normal distribution.
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colline

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« Reply #18140 on: September 23, 2019, 08:06:11 pm »
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Just out of sheer curiosity, what would happen if your CAS dies early on during the exam?

I can't say with 100% certainty, but I'm fairly sure that VCAA expects you to be responsible for bringing a functional and charged CAS to the exam so they won't have any spares available on the day, especially as there are multiple different approved calculators so they can't possibly accommodate for everyone.

Also I couldn't find anything on this page (approved tech for VCAA maths exams) so I would assume there's no spare calculators available.
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« Reply #18141 on: September 23, 2019, 08:27:16 pm »
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Just out of sheer curiosity, what would happen if your CAS dies early on during the exam?
Probably nothing - would need to do everything by hand.

However, something you could do if you have two CAS's is to give one to the assessors before the start of the exam and ask for it if your CAS dies. Obviously would be very assessor dependent on how nice they are. Keep in mind you can only take in one CAS calc.

Also you could take a scientific calculator alongside your CAS as back up for small calculations - obviously wouldn't be able to do any calculus.

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« Reply #18142 on: September 27, 2019, 10:50:19 am »
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VCAA 2007 Q21:

{x: cos^2(x) + 2cos (x) = 0} =

Can someone please explain why the answer is : {x : cos (x) = 0}

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« Reply #18143 on: September 27, 2019, 11:15:05 am »
+1
VCAA 2007 Q21:

{x: cos^2(x) + 2cos (x) = 0} =
Can someone please explain why the answer is : {x : cos (x) = 0}

$(\cos(x))^2 + 2\cos (x) =0 \implies \cos(x)=\frac{-2\pm \sqrt{4-4(1)(0)}}{2(1)} = \frac{-2\pm2}{2}=-2,0.$

But $\cos(x)\ne -2 \implies \cos(x)=0$. Where the quadratic formula was used above to find a solution for $\cos(x)$.
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DrDusk

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« Reply #18144 on: September 27, 2019, 01:56:45 pm »
+1
$(\cos(x))^2 + 2\cos (x) =0 \implies \cos(x)=\frac{-2\pm \sqrt{4-4(1)(0)}}{2(1)} = \frac{-2\pm2}{2}=-2,0.$

But $\cos(x)\ne -2 \implies \cos(x)=0$. Where the quadratic formula was used above to find a solution for $\cos(x)$.
Why are you using the quadratic formula? when you can literally just do

$\cos^2(x) + 2\cos(x) = 0\Rightarrow \cos(x)(\cos(x)+2) = 0$
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Tau

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« Reply #18145 on: September 27, 2019, 01:59:24 pm »
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Why are you using the quadratic formula? when you can literally just do

$\cos^2(x) + 2\cos(x) = 0\Rightarrow \cos(x)(\cos(x)+2) = 0$

That’s fair, works faster and less error prone to factorise. Regardless, same answer.
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Just another student

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« Reply #18146 on: September 27, 2019, 02:02:25 pm »
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Why are you using the quadratic formula? when you can literally just do

$\cos^2(x) + 2\cos(x) = 0\Rightarrow \cos(x)(\cos(x)+2) = 0$

Thanks for the reply, how did simply the equation? like go from first to second step and get rid of the squared?
Also thanks Tau for your approach too!

aspiringantelope

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« Reply #18147 on: September 27, 2019, 02:23:19 pm »
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Thanks for the reply, how did simply the equation? like go from first to second step and get rid of the squared?
Also thanks Tau for your approach too!
Take cos(x) out
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« Reply #18148 on: September 27, 2019, 02:37:48 pm »
+7
It's basic laws of algebra that you learn in year 7 or so?

I don't get what part of that your confused at and why? It's really trivial
Hey, moderator here. This is not an appropriate comment.

You may understand it, but by no means is it okay to label someone's confusion over a subject you find simple 'trivial'.

Please respect others and their individual challenges on this forum.
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« Reply #18149 on: September 27, 2019, 02:43:18 pm »
+2
Hey, moderator here. This is not an appropriate comment.

You may understand it, but by no means is it okay to label someone's confusion over a subject you find simple 'trivial'.

Please respect others and their individual challenges on this forum.
My bad. Upon further inspection I do realize it is being harsh.

Sorry mate @Just another student, I mean no harm =)
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