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November 01, 2025, 10:15:27 am

Author Topic: i dont agree with the solutions.  (Read 2578 times)  Share 

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sam0001

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i dont agree with the solutions.
« on: November 01, 2012, 09:14:02 pm »
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what are you guys getting for this answer?

oneialex

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Re: i dont agree with the solutions.
« Reply #1 on: November 01, 2012, 09:17:16 pm »
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D, from solving for the total number keeping their votes with the same party (400).
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sam0001

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Re: i dont agree with the solutions.
« Reply #2 on: November 01, 2012, 09:18:37 pm »
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Me too. But the answer is c?

dfgjgddjidfg

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Re: i dont agree with the solutions.
« Reply #3 on: November 01, 2012, 09:19:21 pm »
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which exam is this

sam0001

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Re: i dont agree with the solutions.
« Reply #4 on: November 01, 2012, 09:19:58 pm »
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Mav 2012

oneialex

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Re: i dont agree with the solutions.
« Reply #5 on: November 01, 2012, 09:22:58 pm »
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Me too. But the answer is c?
Must be missing something important from the question. I assumed that the order of the elements in each row and column go from Labour to Liberal to Other. Perhaps this isn't necesarily the case, or isn't the way that the question should be viewed?
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dfgjgddjidfg

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Re: i dont agree with the solutions.
« Reply #6 on: November 01, 2012, 09:24:41 pm »
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in the worked solutions they uses matrix mutiplication

oneialex

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Re: i dont agree with the solutions.
« Reply #7 on: November 01, 2012, 09:26:20 pm »
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in the worked solutions they uses matrix mutiplication
Ah, makes sense. It's the logical thing to do. My bad.
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Will T

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Re: i dont agree with the solutions.
« Reply #8 on: November 01, 2012, 09:33:27 pm »
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I got the same answer as oneialex....

It's 200? And it doesn't seem to matter how you order it. (I've tried Labour, Liberal, Other and Liberal, Labour, Other and it gets the same answer, a definite 200). I've also drawn it as a transition network (3 vertices), and I am still getting 200.
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oneialex

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Re: i dont agree with the solutions.
« Reply #9 on: November 01, 2012, 09:41:40 pm »
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I got the same answer as oneialex....

It's 200? And it doesn't seem to matter how you order it. (I've tried Labour, Liberal, Other and Liberal, Labour, Other and it gets the same answer, a definite 200). I've also drawn it as a transition network (3 vertices), and I am still getting 200.
I tried multiplying the given matrix with a 3x1 matrix containing the values 300, 200 and 100. That gave the values 210, 250 and 140. I added the difference between the new values and the old values (90+50+40), which gave 180. It was more-or-less experimentation on my part, I must admit though.
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TheRajinator

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Re: i dont agree with the solutions.
« Reply #10 on: November 01, 2012, 09:43:46 pm »
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wait how are you solving this question? I multiplied the transition matrix with the column matrix and got 210, 250, 140. Where would we go from there?
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oneialex

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Re: i dont agree with the solutions.
« Reply #11 on: November 01, 2012, 09:46:13 pm »
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wait how are you solving this question? I multiplied the transition matrix with the column matrix and got 210, 250, 140. Where would we go from there?
To clear up what I said from my above post, I checked the differences between the new values and the original values (e.g. 300-210=90, 250-200=50, 140-100=40), then added them to give 180. I'm assuming that method is fine? I'm not 100% sure on it.
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Will T

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Re: i dont agree with the solutions.
« Reply #12 on: November 01, 2012, 09:48:24 pm »
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wait how are you solving this question? I multiplied the transition matrix with the column matrix and got 210, 250, 140. Where would we go from there?
To clear up what I said from my above post, I checked the differences between the new values and the original values (e.g. 300-210=90, 250-200=50, 140-100=40), then added them to give 180. I'm assuming that method is fine? I'm not 100% sure on it.

I just tried that with a 3 x 1 matrix containing elements 200;300;100 respectively. I added the differences between those and the original figures and I got 80. The matrix product was 160;310;130. So 40 + 10 + 30 = 80. Therefore, I think we need a different method, as the question doesn't specify which parties are migrating to which parties and this result is incongruent with yours and there would be 4 other ways of ordering the numbers 300, 200, 100 and I imagine some would also have values that are incongruent to 180.
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oneialex

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Re: i dont agree with the solutions.
« Reply #13 on: November 01, 2012, 09:50:39 pm »
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wait how are you solving this question? I multiplied the transition matrix with the column matrix and got 210, 250, 140. Where would we go from there?
To clear up what I said from my above post, I checked the differences between the new values and the original values (e.g. 300-210=90, 250-200=50, 140-100=40), then added them to give 180. I'm assuming that method is fine? I'm not 100% sure on it.

I just tried that with a 3 x 1 matrix containing elements 200;300;100 respectively. I added the differences between those and the original figures and I got 80. The matrix product was 160;310;130. So 40 + 10 + 30 = 80. Therefore, I think we need a different method, as the question doesn't specify which parties are migrating to which parties and this result is incongruent with yours and there would be 4 other ways of ordering the numbers 300, 200, 100 and I imagine some would also have values that are incongruent to 180.
Fair call on that one.  :)
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astone788

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Re: i dont agree with the solutions.
« Reply #14 on: November 01, 2012, 09:51:15 pm »
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easy stuff. Help me with TriG some1!!