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Thread: math help?

  1. 03-10-2009, 12:37 PM #41
    Atlblkz06
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    Are you supposed to use algebra or linear algebra to solve this?
    Use 1000, 10000 and 900 and you'll get pretty close. Tweak it and you have the answer.

    Markov chains - steady state vectors... god damn it, I don't remember how to do this. Look it up online? lol
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  2. 03-10-2009, 01:02 PM #42
    integraboi88
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    Just punch in the numbers on your TI83 plus calculator(problem solved).
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  3. 03-10-2009, 01:28 PM #43
    VWGirl
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    Quote Originally Posted by Atlblkz06
    Are you supposed to use algebra or linear algebra to solve this?
    Use 1000, 10000 and 900 and you'll get pretty close. Tweak it and you have the answer.

    Markov chains - steady state vectors... god damn it, I don't remember how to do this. Look it up online? lol
    Why would you plug those numbers in? It's not a guessing game... and I would not call those numbers close anyway...
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  4. 03-10-2009, 01:35 PM #44
    staysideways
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    Quote Originally Posted by VWGirl
    grades? I'm like a 5th or 6th year senior with 140+ cr hrs
    all those hours and you don't know what year you are?

    i was close to figuring out the answer but i gave up because i have to go to work.

    sorry broseph
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  5. 03-10-2009, 01:44 PM #45
    GLulic
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    ohh you stole the sammich mans girl..lol i wonder what he'll do
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  6. 03-10-2009, 02:05 PM #46
    Atlblkz06
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    Quote Originally Posted by VWGirl
    Why would you plug those numbers in? It's not a guessing game... and I would not call those numbers close anyway...
    I agree, its not a guessing game - I just wanted to give him some numbers close to the answer so he could understand that there is only one combination that can satisfy both since one of the prices is constant.

    Using x= 1k, y=10k, z = 900:
    .25x + 1y + .5z = 10700 (correct)
    .35x + 1.5y + .5z = 15800 (off by $20)

    How is that not close?
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  7. 03-10-2009, 02:18 PM #47
    Princess12
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    A little rusty, but this is my opinion.

    .25x+1y+.50z= 10700
    .35x+1.50y+.50z=15820
    _________________________
    .60x+2.5y+1z=26520

    I would get rid of decimals by multiplying by 10
    10 (.6x+2.5y+z)=26520(10)
    6x+25y+10z=265200

    Now, solve for each variable so you are using common factors.
    x=(265200-25y-10z)/6

    ((265200-25y-10z)/6))+25y-10z=265200

    etc... It will work out this way.

    OR
    Plug the factors into the matrix function on your calculator.
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  8. 03-10-2009, 02:18 PM #48
    keevo54
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    Quote Originally Posted by VWGirl
    so with keevo's help... you should have

    .25x+1y+.50z= 10700
    .35x+1.50y+.50z=15820
    x+y+z=12000

    right?
    u can use a calculator and do it with matrix or...

    x=12000-y-z.......substitute to the first two equations
    .25(12000-y-z)+1y+.5z=10700
    .35(12000-y-z)+1.5y+.5z=15820

    multiply it out and now u have 2 variables and can solve easy...once u have y and z.......u can plug it back in for x
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  9. 03-10-2009, 02:24 PM #49
    VWGirl
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    anyway... i gave him the answers last night cause i was tired and he wasn't getting it... so he has the right answers... it's his own fault when he doesn't understand how to do it to pass a test or move on to the next course
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  10. 03-10-2009, 02:33 PM #50
    Atlblkz06
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    Can someone tell me how to use markov chains to solve this?
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  11. 03-10-2009, 09:21 PM #51
    VWGirl
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    Quote Originally Posted by staysideways
    all those hours and you don't know what year you are?

    i was close to figuring out the answer but i gave up because i have to go to work.

    sorry broseph
    well... for my current BA degree I'd be a senior... but that's two associates degrees and then some... and three different schools in two different states...

    so really i will have enough credit hours by the time i get my BA to have my MA...
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