VCE General & Further Maths Question Thread!

[ayesha2011t]

If you have outliers, your outliers will be beyond your tails.

However, in some circumstances, you may be given a diagram of a box plot where you have single points beyond your tails. In this case, you cannot just assume that they are outliers. Here, you must apply the upper and lower fence formula (Upper Fence: Q3+1.5*IQR and Lower Fence Q1-1.5*IQR) to work out if they are outliers.

Thanks!
Also, how can the standard deviation be used when comparing data distributions in terms of centre and spread?

[clarke54321]

Thanks!
Also, how can the standard deviation be used when comparing data distributions in terms of centre and spread?

If one data distribution has a higher standard deviation than the other, you could conclude that this distribution is more variable than the other. This is because it will be moving further and further away from the mean.

In terms of centre, mean would be a more appropriate measure as opposed to standard deviation.

[ayesha2011t]

If one data distribution has a higher standard deviation than the other, you could conclude that this distribution is more variable than the other. This is because it will be moving further and further away from the mean.

In terms of centre, mean would be a more appropriate measure as opposed to standard deviation.

Thank you!!

[TooLazy]

Hi guys,
I have a question related to time plot graphs. My textbook doesn't seem to explain how to go about with identifying different trends, could someone please explain how to determine whether a graph is cyclic or seasonal. What are the differences and how do you know when either is present in a graph?

Cheers 

[Maths Forever]

Does anyone know if this question involves an inverse relationship? Itute solutions gave a negative answer in their solutions...... This is from the sample exam for the new Further Maths study design.


Question 5 (4 marks)
There is an association between the variables population density, in people per square kilometre,
and area, in square kilometres, of 38 inner suburbs of the same city.
For this association, r2 = 0.141

a. Write down the value of the correlation coefficient for this association between the variables
population density and area.
Round your answer to three decimal places.

Thanks! 

[Maths Forever]

The solutions from Itute:

Q5a
population density = population / area,

inverse relationship between population density and area, .: r = − sqrt(0.141), approximately − 0.375

[TooLazy]

The solutions from Itute:

Does anyone know if this question involves an inverse relationship? Itute solutions gave a negative answer in their solutions...... This is from the sample exam for the new Further Maths study design.


Question 5 (4 marks)
There is an association between the variables population density, in people per square kilometre,
and area, in square kilometres, of 38 inner suburbs of the same city.
For this association, r2 = 0.141

a. Write down the value of the correlation coefficient for this association between the variables
population density and area.
Round your answer to three decimal places.

Thanks! 

Q5a
population density = population / area,

inverse relationship between population density and area, .: r = − sqrt(0.141), approximately − 0.375

Im assuming r2 refers to the correlation coefficient of determination. In this case the correlation coefficient will be the square root of the r2 value. It is only negative if the scatter plot has a negative skew. A negative skew would look like a linear function with a negative gradient.

[clarke54321]

Q5a
population density = population / area,

inverse relationship between population density and area, .: r = − sqrt(0.141), approximately − 0.375


Im assuming r2 refers to the correlation coefficient of determination. In this case the correlation coefficient will be the square root of the r2 value. It is only negative if the scatter plot has a negative skew. A negative skew would look like a linear function with a negative gradient.

Hey,

For this question, however, there is no indication as to whether the correlation is positive or negative.

[Coffee]

A bit of a stupid question I guess but in relation to the attached question, my understanding is that the 6 wouldn't be counted because its heading from the sink to the source and hence would be: 8+4+11=23
But its multiple choice and the options are: 4, 11, 14, 21, and 29.

So, in all cases should I definitely be counting the 6 thereby getting 29?

[clarke54321]

A bit of a stupid question I guess but in relation to the attached question, my understanding is that the 6 wouldn't be counted because its heading from the sink to the source and hence would be: 8+4+11=23
But its multiple choice and the options are: 4, 11, 14, 21, and 29.

So, in all cases should I definitely be counting the 6 thereby getting 29?

Hi Coffee,

Yes, you should be counting the 6 to obtain 29.

[Coffee]

Hi Coffee,

Yes, you should be counting the 6 to obtain 29.

Thanks, Clarke. I was a bit confused since in the textbook it said not to if it was moving upwards like in the above example.

[clarke54321]

Thanks, Clarke. I was a bit confused since in the textbook it said not to if it was moving upwards like in the above example.

Hi Coffee,

As seen in the attached image, the weight on the edge F to B (6) wouldn't be counted as you've already been trapped by the 20. That's the only way I can think of these cuts. 

I agree, that some textbooks can be confusing in their explanations.

[Coffee]

Hope you guys don't mind me asking another question (this one is a bit lengthy). I'm mostly confused by the wording of the question and what its actually asking me to do.

1. The Houndsworth Town Sports Association is planning a sports carnival. Traffic through the town will be diverted around the sporting venues during the carnival. The network (attached below) shows the road that can be used during the carnival represented by the edges and intersection of those roads shown as vertices.
The number on the edges are the maximum number of cars that can travel on the road each hour. The roads will be restricted to allow one-way traffic only, as shown by the arrows.

a. How many cars per hour can enter the diversion roads?
I'm guessing this is just asking from A - B, thus 180? But correct me if I'm wrong.

One cut (Cut 1) is shown on the diagram above.
b. i. If the capacity of Cut 1 is 150, what is the value of m?

ii. Show that the smallest value of m that will ensure there will be no build-up of cars at intersection B is 50.

The capacity of the road between intersection B and intersection E is 50 (m = 50). The maximum flow through the network is currently 150 cars per hour.
c. Mark in the cut that determines the maximum flow on the network above.

I don't quite understand how m = 50? I thought it would just be 150-130=20.

The Sport Association have noticed that there could be a build-up of traffic at intersection E and F.
d. If the road between intersections F and H was improved, what is the maximum capacity is should take to the solve the potential traffic build-up at F?

In order to solve the potential build-up of traffic at intersection E, the Sports Association open another road from intersection E to intersection D.
e. i. What is the minimum capacity of this road that will avoid traffic build-up at intersection E?

ii. Explain why a road between intersection E and D cannot solve potential traffic build-up in the network.

Super sorry for the incredibly lengthy question, I'm just really confused about what the hell its asking me to do and how to go about getting there.

[Kimahso]

Hope you guys don't mind me asking another question (this one is a bit lengthy). I'm mostly confused by the wording of the question and what its actually asking me to do.

1. The Houndsworth Town Sports Association is planning a sports carnival. Traffic through the town will be diverted around the sporting venues during the carnival. The network (attached below) shows the road that can be used during the carnival represented by the edges and intersection of those roads shown as vertices.
The number on the edges are the maximum number of cars that can travel on the road each hour. The roads will be restricted to allow one-way traffic only, as shown by the arrows.

a. How many cars per hour can enter the diversion roads?
I'm guessing this is just asking from A - B, thus 180? But correct me if I'm wrong.

One cut (Cut 1) is shown on the diagram above.
b. i. If the capacity of Cut 1 is 150, what is the value of m?

ii. Show that the smallest value of m that will ensure there will be no build-up of cars at intersection B is 50.

The capacity of the road between intersection B and intersection E is 50 (m = 50). The maximum flow through the network is currently 150 cars per hour.
c. Mark in the cut that determines the maximum flow on the network above.

I don't quite understand how m = 50? I thought it would just be 150-130=20.

The Sport Association have noticed that there could be a build-up of traffic at intersection E and F.
d. If the road between intersections F and H was improved, what is the maximum capacity is should take to the solve the potential traffic build-up at F?

In order to solve the potential build-up of traffic at intersection E, the Sports Association open another road from intersection E to intersection D.
e. i. What is the minimum capacity of this road that will avoid traffic build-up at intersection E?

ii. Explain why a road between intersection E and D cannot solve potential traffic build-up in the network.

Super sorry for the incredibly lengthy question, I'm just really confused about what the hell its asking me to do and how to go about getting there.

1. a. I think you're right but not 100% sure.
b. i. You're right, the answer is 30
ii. The inflow to node B is 180 which comes from the source, which means to prevent congestion the outflow cannot exceed 180. We already have 90+40=130 so m has to be maximum of 50 so that the cut doesn't exceed 180.
c. The cut goes through 30, 40 and 80.
d. The inflow is 100 and the outflow is only 80 which is 20 less. So we should increase it to 100.
e. i. The inflow is 50 and the outflow is only 30 which is 20 less. So the new road from E to D should be at least 20.
ii. I think it's because there's still some congestion at other intersections in the network, for example the outflow of C is 100 which exceeds its inflow of 90. Not 100% sure.

Hope I helped.

[Coffee]

1. a. I think you're right but not 100% sure.
b. i. You're right, the answer is 30
ii. The inflow to node B is 180 which comes from the source, which means to prevent congestion the outflow cannot exceed 180. We already have 90+40=130 so m has to be maximum of 50 so that the cut doesn't exceed 180.
c. The cut goes through 30, 40 and 80.
d. The inflow is 100 and the outflow is only 80 which is 20 less. So we should increase it to 100.
e. i. The inflow is 50 and the outflow is only 30 which is 20 less. So the new road from E to D should be at least 20.
ii. I think it's because there's still some congestion at other intersections in the network, for example the outflow of C is 100 which exceeds its inflow of 90. Not 100% sure.

Hope I helped.

You're brilliant, Kimahso. Thank you so much, that's incredibly helpful!