VCE General & Further Maths Question Thread!

[epichedgehog]

For 4ei, you take your answers from the last part (4d/4c) and apply the rules they give you in the question. ($60 per powered, $30 per unpowered)
The cost function that comes out of that is:
cost = 60x + 30y

Then we just sub in our two minimums from 4d to find the lower cost;
(2,9) cost = 60(2) + 30(9) = $390
(3,8) cost = 60(3) + 30( = $420

So clearly (2,9) is the real minimum of the function, with the lowest cost being $390

For 4eii,
I'm not 100% sure on the logic, but here's itute's solutions which are a bit clearer than the VCAA's.

Thank you very much, I understand now .

[epichedgehog]

For this question, you'd obviously have to refer to the solution region in the graph, you can see that there are some integer points that could be solutions in your graph right?

Since in this context we're talking about an integer number of things (power and unpowered camps) we really can't have the maximum at the corner vertex, since we're more than likely to get a decimal amount of camp sites, which is impossible.

By looking at the solution region, you can see the smallest integer number of powered and unpowered camp sites will occur at either 2,9 or 3,8. Anything beyond below that is a decimal or not in the solution region so it has to be one of those two points. using the objective function 60x+30y you find that the point 2,9 gives the minimum cost which is $390.

For question 4 eii you have to understand that since all girls have to use one campsite and all boys have to use one campsites,  it means that there are 24 boys at powered campsite and 24 girls and unpowered camp sites. powered campsites carry 6 people, 24/6 = 4, so X must equal to 4 powered campsites required is = 4. Unpowered campsites carry 4 people, 24/4 = 6.  unpowered campsites required would be 6.

So look at the solution region, you see that 4,6 isn't in that region and it has to match constraints y => 2x, to match this constraint y= 2* 4 and so y must = 8. so the points we have now are 4,8

Now just find the cost using the objective function 60x+30y, and you get 480$.



It's a bit long but i hope you understand..


EDIT: So MightyBeh beat me to it

thanks! I was just a little confused about why x = 4 and y = 6, but now it's much clearer.

[n.a]

How do you tell the difference between cyclic and random variations in time series? I understand seasonal variation has to have regular intervals between the peaks and troughs, but is it possible for it to get confusing with cyclic sometimes as well?

[ineedbiohelp]

How do you tell the difference between cyclic and random variations in time series? I understand seasonal variation has to have regular intervals between the peaks and troughs, but is it possible for it to get confusing with cyclic sometimes as well?

From everything ive seen, theres been no confusing ones,
However, these videos might help you as she explains it good, https://www.youtube.com/watch?v=ca0rDWo7IpI
At six minutes she explains what youre talking about.

[ineedbiohelp]

Hello again guys,
Heres a matrices  question aswell as a core one..
For the core one i have no idea why the answer gave would best suit the stem and leaf plot, as i believe distance travelled would be a better answer...

For the matrices one, i have only seen this question once, and im confused on how to do it. If anyone has any idea on the theory behind it, or atleast the method of solving any question like that, i will be grateful.

[Law]

Hello again guys,
Heres a matrices  question aswell as a core one..
For the core one i have no idea why the answer gave would best suit the stem and leaf plot, as i believe distance travelled would be a better answer...

For the matrices one, i have only seen this question once, and im confused on how to do it. If anyone has any idea on the theory behind it, or atleast the method of solving any question like that, i will be grateful.

Core question:
A back-to back stem plot measures bivariate data, involving a numerical variable and a categorical variable with 2 categories. Thus it would car's speed (numerical value) and the only categorical variable with 2 categories (drivers sex: male or female)

Matrices Question:
So T1 x S0= S1. To make S1 and S2 constant you need to add and subtract the number that changes from S1 to S2.

There is an illustration in the attachment.

Edit: Raw 50 paint job skills

[TheMereCat]

Hello again guys,
Heres a matrices  question aswell as a core one..
For the core one i have no idea why the answer gave would best suit the stem and leaf plot, as i believe distance travelled would be a better answer...

For the matrices one, i have only seen this question once, and im confused on how to do it. If anyone has any idea on the theory behind it, or atleast the method of solving any question like that, i will be grateful.

For the Core question, the one that make the most sense is answer E, which is Driver's sex male or female. Since you're sort of comparing only two things side by side (Categorical variable) and average speed in km/hr (numerical variable) the best suited would be a back to back stemplot, given that there is 1) a categorical variable and a numerical variable. and 2) Since there is only two categories, Male and female. Distance travelled won't really work since it is a numerical variable as well. A back to back stemplot is designed for a categorical and numerical variable.


For the Matrices one, you want to know how many eggs, baby trout and adult trout, that could be added or removed at the pond of each year will allow the population to remain constant.

For this question you firstly have to find S₁ and once you do that you takeaway S₁ from S₀. Basically S₀-S₁ Once you found that difference, you know that if you add that amount every year, the population will remain constant.


Not great at explaining that one, but if you need further explanation let me know.


EDIT: Law already answered

[ineedbiohelp]

For the Matrices one, you want to know how many eggs, baby trout and adult trout, that could be added or removed at the pond of each year will allow the population to remain constant.

For this question you firstly have to find S₁ and once you do that you takeaway S₁ from S₀. Basically S₀-S₁ Once you found that difference, you know that if you add that amount every year, the population will remain constant.


Not great at explaining that one, but if you need further explanation let me know.


EDIT: Law already answered

Okay i understand the core one, thanks to both of you,

But for the matrices one, what does it mean  for the population to remain constant? i assume it means like a steady state so the values wont change at all from one year to the next, but when i subsitute these new added values into the S0, the values change yet again? Im pretty sure i can just always use S0-S1 if the question comes up, but id like to understand what it means as ill understand it easier,

Thanks

[Law]

Okay i understand the core one, thanks to both of you,

But for the matrices one, what does it mean  for the population to remain constant? i assume it means like a steady state so the values wont change at all from one year to the next, but when i subsitute these new added values into the S0, the values change yet again? Im pretty sure i can just always use S0-S1 if the question comes up, but id like to understand what it means as ill understand it easier,

Thanks

Dont add them to S0 because thats the initial state, you cant change it. However the question asks at the end of each year. So you have to change S1. Remenber T1 x S0= S1.

For example= Lets take Eggs. At S0 Eggs=10,000. Then at the end of the year it equals 0. So to make it constant it each year, to make it equal 0 each year, you would need to add 10,000 at the end of each year. Thus S1=S2 which =S3.

[Orson]

How do you guys do those weird matrices questions? The ones with m and n and whatnot. I always find myself guessing them in the end. (Usually MCQs).

Thanks!

[bedigursimran]

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.

Hey. I found a worked solution for an insight exam 1 for a transformed data (as you can see in the attached image) they just substituted in points instead of finding the gradient. So confused about this! Please help anyone.

[keltingmeith]

How do you guys do those weird matrices questions? The ones with m and n and whatnot. I always find myself guessing them in the end. (Usually MCQs).

Thanks!

Having no idea what type of questions you're talking about, my suggestion is to sub in values for n and m and see if that helps you at all. (I might be able to give better advice if you show me a question, hahah.)

[bedigursimran]

How do you guys do those weird matrices questions? The ones with m and n and whatnot. I always find myself guessing them in the end. (Usually MCQs).

Thanks!

What I do is I put the letters on my calculator and it usually works out.

[Law]

Hey. I found a worked solution for an insight exam 1 for a transformed data (as you can see in the attached image) they just substituted in points instead of finding the gradient. So confused about this! Please help anyone.

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.

[Law]

How do you guys do those weird matrices questions? The ones with m and n and whatnot. I always find myself guessing them in the end. (Usually MCQs).

Thanks!

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