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  1. #1
    Ola
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    maths Help

    I have this done half way but not sure of how the remaining should be handled, pleaes help me out.
    The equationP = P0 (1-e-0.05T) has been developed to simulate the spread of a virus and relates the number of persons infected (P) in a population P0 , after a period of T days after it has first been detected.
    Use this model to determine:
    (i) The number of persons in a town with a population of 10,000 that would be expected to be infected five days after the first detection of the virus
    (ii) The time in terms of number of days before 50% of the population is infected
    (iii) The time in days before 95% of the population is infected.
    Here is my steps:
    P = 10,000 (1 e-0.0(5)(5))
    P = 10,000 (1 e-0.025)
    P = 10,000 (1- 1/e0.025)

    I was stocked at how to convert 1/e0.025.

    Please bail me out.

    Ola
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  2. #2
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    Quote Originally Posted by Ola View Post
    I have this done half way but not sure of how the remaining should be handled, pleaes help me out.
    The equationP = P0 (1-e-0.05T) has been developed to simulate the spread of a virus and relates the number of persons infected (P) in a population P0 , after a period of T days after it has first been detected.
    Use this model to determine:
    (i) The number of persons in a town with a population of 10,000 that would be expected to be infected five days after the first detection of the virus
    (ii) The time in terms of number of days before 50% of the population is infected
    (iii) The time in days before 95% of the population is infected.
    Here is my steps:
    P = 10,000 (1 e-0.0(5)(5))
    P = 10,000 (1 e-0.025)
    P = 10,000 (1- 1/e0.025)

    I was stocked at how to convert 1/e0.025.

    Please bail me out.

    Ola
    1. I assume that the given equation reads:

    P(T)=P_0\left(1-e^{-0.05 \cdot T}  \right)

    If so then

    (i) would be: P(5)=10,000 \left(1-e^{-0.05 \cdot 5}  \right) = 10,000 \left(1-e^{-0.25}  \right)<br />
    and now use your calculator. Keep in mind that there aren't any decimal fractions of persons.

    (ii) would be: \frac12 P_0 = P_0  \left(1-e^{-0.05 \cdot T}  \right)

    Solve for T.

    (iii) should be done in exactly the same way as (ii).
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  3. #3
    Ola
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    Maths - Exponential

    Hi,

    Very grateful for your help, pls can you help me to have a look at the solutions I finally arrived at just to be sure I am on the right track:

    (i)
    P=Po(1-e^-0.05(5))
    P = 10,000(1-^-0.25)
    P = 10000 (1-0.7788)
    P = 10,000(0.2212)
    P = 2,212

    (ii)
    1/2 of 10,000 (1-e^0.05(T))
    = 1/2(10,000) (1 - 0.95123T)
    = 5,000 (1 - 0.95123T
    = 5,000 - 4756.15T
    4756.15T = 5,000
    T = 5,000/4756.15
    T = 1.05127
    T = 1 day (aproximately)

    (iii)
    95% of 10,000 (1 - e^0.05(T))
    = 9,500 (1 - 0.95123T)
    = 9,500 - 9,036.68T
    9036.68T = 9,500
    T = 9,500/9036.68
    T = 1.051271 day
    T = 1 day (aproximately)
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  4. #4
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    Quote Originally Posted by Ola View Post
    Hi,

    Very grateful for your help, pls can you help me to have a look at the solutions I finally arrived at just to be sure I am on the right track:

    (i)
    P=Po(1-e^-0.05(5))
    P = 10,000(1-^-0.25)
    P = 10000 (1-0.7788)
    P = 10,000(0.2212)
    P = 2,212

    (ii)
    1/2 of 10,000 (1-e^0.05(T))
    = 1/2(10,000) (1 - 0.95123T)
    = 5,000 (1 - 0.95123T
    = 5,000 - 4756.15T
    4756.15T = 5,000
    T = 5,000/4756.15
    T = 1.05127
    T = 1 day (aproximately) <<<<< if you have 2000 infections in 5 days it would take a little bit longer to have 5000 infections

    (iii)
    95% of 10,000 (1 - e^0.05(T))
    = 9,500 (1 - 0.95123T)
    = 9,500 - 9,036.68T
    9036.68T = 9,500
    T = 9,500/9036.68
    T = 1.051271 day
    T = 1 day (aproximately) <<<<<<< No
    to (ii):

    50% of 10,000 = 5,000 . Plug in this value for P(T) and solve for T:

    5000 = 10000 \left(1-e^{-0.05 \cdot T} \right)~\implies~\dfrac12= 1-e^{-0.05 \cdot T}~\implies~\dfrac12= e^{-0.05 \cdot T}<br />

    Now use logarithms to solve for T.

    \ln\left(\frac12\right)=-0.05\cdot T~\implies~T=\dfrac{\ln\left(\frac12\right)}{0.05}  \approx 13.86\ days

    (iii) has to be done in exactly the same way.
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  5. #5
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    I accept with information:I have this done half way but not sure of how the remaining should be handled, pleaes help me out.
    The equationP = P0 (1-e-0.05T) has been developed to simulate the spread of a virus and relates the number of persons infected (P) in a population P0 , after a period of T days after it has first been detected.
    Use this model to determine:
    (i) The number of persons in a town with a population of 10,000 that would be expected to be infected five days after the first detection of the virus
    (ii) The time in terms of number of days before 50% of the population is infected
    (iii) The time in days before 95% of the population is infected.
    Here is my steps:
    P = 10,000 (1 e-0.0(5)(5))
    P = 10,000 (1 e-0.025)
    P = 10,000 (1- 1/e0.025)

    I was stocked at how to convert 1/e0.025.


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  6. #6
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    Quote Originally Posted by cinlymcre View Post
    ...

    I was stocked at how to convert 1/e0.025.

    1. I don't know why you want to do this ...

    2. According to my first reply the term should be: 1-e^{-0.25}

    3. Since 0.25 = \dfrac14 the term becomes:

    1-e^{-\frac14} = 1-\dfrac1{e^{\frac14}} = 1-\dfrac1{\sqrt[4]{e}}
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  7. #7
    Ola
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    Hi,
    I know it will be a surprise to you. My background in maths is terrible very sorry for the mistake and that is the reason why I have come to somebody like yourself to rescue me. I appreciate your efforts.
    Ola.
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