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Math Help - problem solving

  1. #1
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    problem solving

    P(t)=0.8/(1+1.67e^-0.16t)
    The model represents the proportion of new computers sold that use windows 98. Let t=0 represent 1998, t=1 represent 1999, and so on.

    1)determine the maximum proportion of new computers sold that will use win 98. That is, determine the limit of P(t) as t -> infinity

    2)using a graphing utlility graph p(t)

    3)When will 75% of new computers sold use win 98?
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  2. #2
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    e^(i*pi)'s Avatar
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    Quote Originally Posted by dmak263 View Post
    P(t)=0.8/(1+1.67e^-0.16t)
    The model represents the proportion of new computers sold that use windows 98. Let t=0 represent 1998, t=1 represent 1999, and so on.

    1)determine the maximum proportion of new computers sold that will use win 98. That is, determine the limit of P(t) as t -> infinity

    2)using a graphing utlility graph p(t)

    3)When will 75% of new computers sold use win 98?
    Someone may want to double check this >.<

    1. Think about what happens to e^x for a large negative x.

    Spoiler:
    \text {as} t \rightarrow \infty , e^{-0.16t} \rightarrow 0

    <br />
P_{\infty} = \frac{0.8}{1} = \frac{4}{5}


    2. Plug in the values and away you go. I'll edit in my graph later

    3. As P(t) = 0.75, solve for t. (Isolate e^-0.16t and take logs)

    Spoiler:
    0.75 =\frac{0.8}{1+1.67e^{-0.16t}}

    Take the reciprocal of both sides:

    \frac{4}{3} = \frac{1+1.67e^{-0.16t}}{0.8}

    0.8\frac{4}{3} = 1+1.67e^{-0.16t}

    \frac{16}{15} - 1 = 1.67e^{-0.16t}

    e^{-0.16t} = \frac{15}{16} \times \frac{1}{1.67} = \frac{375}{668}

    t = -\frac{1}{0.16}ln(\frac{375}{668}) = 3.61 years
    Last edited by e^(i*pi); April 20th 2009 at 08:14 AM. Reason: missed a bit
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  3. #3
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    I don't understand the steps for #1...#3, I comprehend completely.
    There was another question... what proportion of new computers sold in 1998 use windows 98?

    my answer -
    p(0)= 0.8/[1+1.67e^-0.16^(0)]
    =0.8/2.67
    =.299
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