VCE General & Further Maths Question Thread!

[bedigursimran]

Always think about these type of questions as y=(any variable) multiplied by x. For the sake of this question. Lets make this variable (k)

In this question y is plotted against x^3.
The rule connecting y and x would thus be y=kx^3.

We got coordinates in the graph= (8,2)
We sub these points in= 2=k8.

Now when we sub x in= It is not 8^3 its simply 8 because its already cubed. And the rule is y=kx^3

So solve(y=k x 8,k) and k equals= 1/4.

Thus y=1/4 times x^3. We can simplify this to x^3/4.
Thus y=x^3/4.
That is C.

This is what TheMereCat said:

You can't substitute the coordinates to find k here, because it's plotted as H against 1/d^2, if you had the original graph, which would be H against D, you would be able to use it to find the coefficient. Since you have the transformed graph, you'll need to find the K using the gradient of the straight line.

[Law]

This is what TheMereCat said:

For the question you posted I subbed in the co-ordinates and got the right answer. So yeah you can use them.

I see from his answer that he is talking about d^2, so its a different question. I looked for a bit but i cant find it. I have always subbed in the co-ordinates and got all the answers right though, so it might be a different kind of question.

[cipskailok1]

need help with this question from paper 1 VCE 2013 : Module 1 Number Patterns, Question 8.

Question 8

The initial rate of pay for a job is $10 per hour.
A worker’s skill increases the longer she works on this job. As a result, the hourly rate of pay increases each
month.
The hourly rate of pay in the nth month of working on this job is given by the difference equation

Sn+1= 0.2×Sn + 15         S1= 10

The maximum hourly rate of pay that the worker can earn in this job is closest to
A.
  $3.00
B.
 $12.00
C.
 $12.50
D.
 $18.75
E.
 $75.00

[RazzMeTazz]

Is the moving average smoothing technique affected by outliers ? For this reason would it be more advantageous to use median smoothing if outliers are present ? Thanks

[BakedDwarf]

Is the moving average smoothing technique affected by outliers ? For this reason would it be more advantageous to use median smoothing if outliers are present ? Thanks

Yep!

[Orson]

Here is variable matrices question from the past 3 years. Do you want me to post more?

OMG Mr/Ms. Law. That would be amazing if you could do that!

Thanks so much. I really need to do well, after yesterday's fiasco...

[n.a]

Would you ever need to find outliers of bivariate data? If so, how would you do it?

[epichedgehog]

hi all,
here's a question from NEAP 2006 graphs + relations (Q10):
shouldn't b, c and e all be suitable options, according to the circle of transformations? i'm a little confused at NEAP's explanation, i don't remember learning anything like it in class.
thanks!

[TheMereCat]

For Question 9, is the question incorrect?

Since residual = y-ypredicted.

Shouldn't you have to use the equation, then square root the value to find y, then do 2-2.5

to get something close to ~0.5?

apparently the answer is B, and they squared the y value- minus y² predicted value....

[BakedDwarf]

hi all,
here's a question from NEAP 2006 graphs + relations (Q10):
shouldn't b, c and e all be suitable options, according to the circle of transformations? i'm a little confused at NEAP's explanation, i don't remember learning anything like it in class.
thanks!

Yeah, they all should be suitable... so just ignore it

For Question 9, is the question incorrect?

Since residual = y-ypredicted.

Shouldn't you have to use the equation, then square root the value to find y, then do 2-2.5

to get something close to ~0.5?

apparently the answer is B, and they squared the y value- minus y² predicted value....

Yes, this question is correct.

It's asking for the residual for the transformed data. Not the original data

[AngelWings]

1. Good luck for your Exam 1 tomorrow, people. I know you'll all do well.
2. I edited my previous post answering bedigursman's question about VCAA 2013 Exam 1 Matrices Q7 and it now has a detailed explanation on the attachment, just to let anyone who wants help with that question know.
3.

need help with this question from paper 1 VCE 2013 : Module 1 Number Patterns, Question 8.

Question 8

The initial rate of pay for a job is $10 per hour.
A worker’s skill increases the longer she works on this job. As a result, the hourly rate of pay increases each
month.
The hourly rate of pay in the nth month of working on this job is given by the difference equation

Sn+1= 0.2×Sn + 15         S1= 10

The maximum hourly rate of pay that the worker can earn in this job is closest to
A.
  $3.00
B.
 $12.00
C.
 $12.50
D.
 $18.75
E.
 $75.00

Put this sequence into the calculator and check when it starts to become constant.
Take the number at which it plateaus at.
You'll get 18.75.
Answer is D.

Would you ever need to find outliers of bivariate data? If so, how would you do it?

Yeah, but it's more likely to be an Exam 1 type question and unlikely to appear; chances are they'll ask for the residual from a given point with a given equation (see above two posts for an example).

Back to the point, they're generally very clear about it. For example, a scatterplot with a point that's completely away from everything else (hence the method is self explanatory) and they could possibly ask you something about the outlier.
e.g. Not just "Find coordinates of the outlier.", but "Which of the following methods would be affected by an outlier?" and so forth.

[n.a]

Thanks, AngelWings! Super nervous about tomorrow! :/

Another question:
VCAA 2012 Q6 from Geo&Trig:

I got it wrong and understand that it has to do with linear and volume scale factors. But the reason I got it wrong was because I did the following:

V=((pi) x (r^2)) x h
V2=((pi) x (3r^2)) x 3h
This gave me 9( (pi) x (h) x (r^2) ).
Then I just subbed in (pi) x (h) x (r^2)=V, and got 9V.

What did I do wrong?

[StupidProdigy]

Thanks, AngelWings! Super nervous about tomorrow! :/

Another question:
VCAA 2012 Q6 from Geo&Trig:

I got it wrong and understand that it has to do with linear and volume scale factors. But the reason I got it wrong was because I did the following:

V=((pi) x (r^2)) x h
V2=((pi) x (3r^2)) x 3h  (3r)^2 not 3(r^2) I think just having a quick look
This gave me 9( (pi) x (h) x (r^2) ).
Then I just subbed in (pi) x (h) x (r^2)=V, and got 9V.

What did I do wrong?

[BakedDwarf]

Thanks, AngelWings! Super nervous about tomorrow! :/

Another question:
VCAA 2012 Q6 from Geo&Trig:

I got it wrong and understand that it has to do with linear and volume scale factors. But the reason I got it wrong was because I did the following:

V=((pi) x (r^2)) x h
V2=((pi) x (3r^2)) x 3h
This gave me 9( (pi) x (h) x (r^2) ).
Then I just subbed in (pi) x (h) x (r^2)=V, and got 9V.

What did I do wrong?

V = pi x R^2 x H

Therefore, because the R is multiplied by 3 and H is multiplied by 3, the volume becomes:

V = pi x (3R)^2 x 3H
   = pi x 9R x 3H
   = 27 x pi x R x H

This is the same as:

27 x V

Hence, the answer is D

EDIT: StupidProdigy identified the mistake

[AngelWings]

Thanks, AngelWings! Super nervous about tomorrow! :/

You guys have worked hard (I assume), so you guys have nothing to worry about. Try your best, breathe, be prepared and smash it out of the park.

While you're on that paper, n.a, have a look at Q9 of the Geometry section.