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October 20, 2025, 04:57:23 am

Author Topic: My Solutions (Core; Graphs; Networks; Matrices)  (Read 29846 times)  Share 

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brenden

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Re: My Solutions (Core; Graphs; Networks; Matrices)
« Reply #60 on: November 02, 2012, 04:52:27 pm »
I'm not debating the answer and I've skimmed read so I don't think anyone else has asked the question...
Question 5, Core, why answer D? I had histogram. I feel good, I ran out of time and randomly shaded 3 answers, one of them was correct, so lost 2 on time, and this Core question I also lost. So looks like 37/40. Bloody hope the cut off is 36.
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Re: My Solutions (Core; Graphs; Networks; Matrices)
« Reply #61 on: November 02, 2012, 04:53:33 pm »
Question 5 networks, you can test it on the second diagram (the pentagon sorta diagram) and an Euler circuit exists. So either way, the answer can't be 0.
I see now. I have found a multitude of Eulerian circuits for the pentagon-shaped complete graph. I will try a similar process for the heptagon and nonagon. This means the range of possible answers is now 1-3 so alternatives B, C or D could be correct at the moment.

Its definitely 3.

On a complete graph with n vertices, the degree of each vertex is (n-1) (since it connects to all other vertices)

The graphs have 4,5,7,9 vertices respectively.

for n-1 to be even, n has to be odd.
For a Eulerian circuit to exist, degree of every vertex has to be even.

So Eulerian circuits exist in K5,7,9 so the answer is 3

Jezza

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Re: My Solutions (Core; Graphs; Networks; Matrices)
« Reply #62 on: November 02, 2012, 04:54:27 pm »
I'm not debating the answer and I've skimmed read so I don't think anyone else has asked the question...
Question 5, Core, why answer D? I had histogram. I feel good, I ran out of time and randomly shaded 3 answers, one of them was correct, so lost 2 on time, and this Core question I also lost. So looks like 37/40. Bloody hope the cut off is 36.
Yeah same here. I went with C - Histogram, but it makes sense that it's time series plot, because it's obviously talking about the time of the temperature.

Yendall

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Re: My Solutions (Core; Graphs; Networks; Matrices)
« Reply #63 on: November 02, 2012, 04:55:00 pm »
I'm not debating the answer and I've skimmed read so I don't think anyone else has asked the question...
Question 5, Core, why answer D? I had histogram. I feel good, I ran out of time and randomly shaded 3 answers, one of them was correct, so lost 2 on time, and this Core question I also lost. So looks like 37/40. Bloody hope the cut off is 36.
I had Histogram as well. Apparently time series is more suitable, but I think both would work fine.

I saw it as two sets of numeric data, and the list of possibilities was:
  • Histograms
  • Frequency polygons
  • Cumulative frequency polygons
  • Boxplots
  • Stemplots
  • Dotplots
I didn't think Time Series Plots would even come into the equation when summarising two sets of Numerical Data.
« Last Edit: November 02, 2012, 04:57:42 pm by Yendall »
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oneoneoneone

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Re: My Solutions (Core; Graphs; Networks; Matrices)
« Reply #64 on: November 02, 2012, 04:59:20 pm »
I'm not debating the answer and I've skimmed read so I don't think anyone else has asked the question...
Question 5, Core, why answer D? I had histogram. I feel good, I ran out of time and randomly shaded 3 answers, one of them was correct, so lost 2 on time, and this Core question I also lost. So looks like 37/40. Bloody hope the cut off is 36.
I had Histogram as well. Apparently time series is more suitable, but I think both would work fine.

I saw it as two sets of numeric data, and the list of possibilities was:
  • Histograms
  • Frequency polygons
  • Cumulative frequency polygons
  • Boxplots
  • Stemplots
  • Dotplots
I didn't think Time Series Plots would even come into the equation when summarising two sets of Numerical Data.

Histograms count frequency/percentage frequency though. You don't count a discrete amount of degrees, its a continuous variable.

oneialex

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Re: My Solutions (Core; Graphs; Networks; Matrices)
« Reply #65 on: November 02, 2012, 05:06:56 pm »
The last question in business maths is C, not A

No, its A. If you put all the information into the calculator, then you end up with a final value of about $2. As this is not 0, it can't be C.

the question states that "he planed to repay the loan fully", however in reality he doesn't fully pay it off.
But as the question doesn't state the original balance of the loan (as far as I can see), isn't it correct to assume that he does repay it in full? I'm quite possibly missing something here, feel free to give me the details if I'm wrong.
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brenden

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Re: My Solutions (Core; Graphs; Networks; Matrices)
« Reply #66 on: November 02, 2012, 05:11:48 pm »
The last question in business maths is C, not A

No, its A. If you put all the information into the calculator, then you end up with a final value of about $2. As this is not 0, it can't be C.

the question states that "he planed to repay the loan fully", however in reality he doesn't fully pay it off.
But as the question doesn't state the original balance of the loan (as far as I can see), isn't it correct to assume that he does repay it in full? I'm quite possibly missing something here, feel free to give me the details if I'm wrong.
Nope, pick a random number and percentage and chuck it on TVM and fiddle with it and you won't get a fully repayed loan. It won't be 0 because 260 equal payments got it to zero, correct? That was consistent, equal payments. When you skip a payment, his principal increases slightly, because nothing has been paid back but the interest rate has increased it a little bit. Thus, even though he double the payment to 520, it does not take into account that slight increase when the fourth payment was missed. Does that make sense?
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oneialex

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Re: My Solutions (Core; Graphs; Networks; Matrices)
« Reply #67 on: November 02, 2012, 05:18:09 pm »
The last question in business maths is C, not A

No, its A. If you put all the information into the calculator, then you end up with a final value of about $2. As this is not 0, it can't be C.

the question states that "he planed to repay the loan fully", however in reality he doesn't fully pay it off.
But as the question doesn't state the original balance of the loan (as far as I can see), isn't it correct to assume that he does repay it in full? I'm quite possibly missing something here, feel free to give me the details if I'm wrong.
Nope, pick a random number and percentage and chuck it on TVM and fiddle with it and you won't get a fully repayed loan. It won't be 0 because 260 equal payments got it to zero, correct? That was consistent, equal payments. When you skip a payment, his principal increases slightly, because nothing has been paid back but the interest rate has increased it a little bit. Thus, even though he double the payment to 520, it does not take into account that slight increase when the fourth payment was missed. Does that make sense?
Unfortunately it does, hahahaha. Thanks for the info  :)
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brenden

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Re: My Solutions (Core; Graphs; Networks; Matrices)
« Reply #68 on: November 02, 2012, 05:19:55 pm »
You're welcome :)
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davidbrock

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Re: My Solutions (Core; Graphs; Networks; Matrices)
« Reply #69 on: November 02, 2012, 05:35:56 pm »
hey how is question 2 of the core, B? isnt it E
« Last Edit: November 02, 2012, 05:38:47 pm by davidbrock »

Will T

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Re: My Solutions (Core; Graphs; Networks; Matrices)
« Reply #70 on: November 02, 2012, 05:40:35 pm »
Question 5 networks, you can test it on the second diagram (the pentagon sorta diagram) and an Euler circuit exists. So either way, the answer can't be 0.
I see now. I have found a multitude of Eulerian circuits for the pentagon-shaped complete graph. I will try a similar process for the heptagon and nonagon. This means the range of possible answers is now 1-3 so alternatives B, C or D could be correct at the moment.

Its definitely 3.

On a complete graph with n vertices, the degree of each vertex is (n-1) (since it connects to all other vertices)

The graphs have 4,5,7,9 vertices respectively.

for n-1 to be even, n has to be odd.
For a Eulerian circuit to exist, degree of every vertex has to be even.

So Eulerian circuits exist in K5,7,9 so the answer is 3

I am in agreement with you now, I just had to check some stuff, thanks.
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Re: My Solutions (Core; Graphs; Networks; Matrices)
« Reply #71 on: November 02, 2012, 05:41:22 pm »
hey how is question 2 of the core, B? isnt it E

No, its a bar chart, not a percentage frequency graph. So its not 35%, but actually 35/214 which is 16%
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davidbrock

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Re: My Solutions (Core; Graphs; Networks; Matrices)
« Reply #72 on: November 02, 2012, 05:47:04 pm »
hey how is question 2 of the core, B? isnt it E

No, its a bar chart, not a percentage frequency graph. So its not 35%, but actually 35/214 which is 16%

god damnit

bro0012

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Re: My Solutions (Core; Graphs; Networks; Matrices)
« Reply #73 on: November 02, 2012, 05:51:07 pm »
I thought histograms were only for one variable? and the other axis being frequency of that variable?

oneoneoneone

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Re: My Solutions (Core; Graphs; Networks; Matrices)
« Reply #74 on: November 02, 2012, 05:52:45 pm »
Bro thats right. If you're talking about the temperature one it should be a Time-Series plot.