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Math Help - exponential model equations

  1. #1
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    exponential model equations

    Before I begin I just want to say anything underlined (e.g. L) is subscript and anyting with ^ infront of it means its a power.

    (A) In 1988 number of reported new cases of males with AIDS in a country was 27,049. In 1989 the number of new cases reported was 29,549 males.

    (i) If the rate of change in the number of new cases stays constant, find a linear function of the form AL(t) = mt + c that would model the number AL of reprted new cases of males with AIDS per year t, where t is the number of years after 1988.

    (ii)If the number of new cases instead reflects a constant percentage increase, find an exponential function of the form AE(t) = A0(a)^t that would similarly model the number of reported new cases of male with AIDS.

    (iii)In 1992 the actual number of reported new cases of males with AIDS was 38917. Which model best predicts the growth in AIDS according to this data?



    (B) In 1990, the number of families living below the poverty line was 33,585,000 families. By 1991, the number had increased 6%.

    (i) Write a function of the form P(t) = P0a^t to model this growth, where P(t) is the population t years after 1990.

    (ii) How many families will be living below the poverty line now (2007) according to this model?

    (iii) When will the number of families living below the poverty line reach 1 billion if the growth continues at 6%?

    Someone please help me...
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  2. #2
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    Quote Originally Posted by Unt0t View Post
    ...

    (A) In 1988 number of reported new cases of males with AIDS in a country was 27,049. In 1989 the number of new cases reported was 29,549 males.

    (i) If the rate of change in the number of new cases stays constant, find a linear function of the form AL(t) = mt + c that would model the number AL of reprted new cases of males with AIDS per year t, where t is the number of years after 1988.

    (ii)If the number of new cases instead reflects a constant percentage increase, find an exponential function of the form AE(t) = A0(a)^t that would similarly model the number of reported new cases of male with AIDS.

    (iii)In 1992 the actual number of reported new cases of males with AIDS was 38917. Which model best predicts the growth in AIDS according to this data?
    ...

    Hello, UntOt,

    to A):
    (i)
    1. the y-intercept of the line is 27049
    2. the slope is a quotient of differences: (29549-27049)/(1989-1988) = 2500

    A_(L)(t) = 27049 + 2500*t

    (ii)
    1. calculate the base a. It is the relative increase per year: 29549/(27049*1 yr) ≈ 1.092425

    A_(E)(t) = 27049*(1.092425)^t

    (iii)

    In 1992 t = 4

    A_(L)(4) = 27049 + 2500*4 = 37049

    A_(E)(4) = 27049*(1.092425)^4 ≈ 38523

    Thus the 2nd equation predicts the increase of new infections better than the first one.
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  3. #3
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    Hey thanks heaps, now can anyone help me out with question B?
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  4. #4
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    Quote Originally Posted by Unt0t View Post
    ...
    (B) In 1990, the number of families living below the poverty line was 33,585,000 families. By 1991, the number had increased 6%.

    (i) Write a function of the form P(t) = P0a^t to model this growth, where P(t) is the population t years after 1990.

    (ii) How many families will be living below the poverty line now (2007) according to this model?

    (iii) When will the number of families living below the poverty line reach 1 billion if the growth continues at 6%?
    ...
    Hello, UntOt,

    (i)
    if the relative increase is 6% then the factor (1 + 0.06) is the base of the function. Thus:
    P(t) = 33,585,000 * (1.06)^t

    (ii) from 1990 to 2007 you have t = 17. Calculate

    P(17) = 33,585,000 * (1.06)^17 ≈ 90,436,774 . Because the initial value is exact to the thousands I would use 90,437,000.

    (iii)

    You know P(t) = 10^9 . Plug in this value into the equation and solve for t:

    1,000,000,000 = 33,585,000 * (1.06)^t . Divide by 33,585,000

    29.7752 = (1,06)^t . Use logarithm

    log(29.7752) = log(1.06) * t
    Code:
         log(29.7752)
    t =  ------------ ≈ 58.24 years
         log(1.06)
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  5. #5
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    Thanks heaps mate, you're a saviour
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  6. #6
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    Quote Originally Posted by Unt0t View Post
    Thanks heaps mate, you're a saviour
    Hi,

    you are welcome !


    I'm a little bit curious about your name "UntOt"... It's a German word meaning "undead" or vulgo "zombie".
    Are you looking like that?
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  7. #7
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    haha i'm not german, but I have a german friend who gave me the name for computer games and such. Kind of stuck with me
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