A town has a population of 6000 people in the year 2009 and is growing at a rate of 3% a year. Let t denote the time in years after 2009, so

Question

A town has a population of 6000 people in the year 2009 and is growing at a rate of 3% a year. Let t denote the time in years after 2009, so t = 0 denotes 2009. Let f be the function that, to each year t, assigns the population of the town in year t. In the text box below, enter a formula for f(t). Your entry must begin with f(t)= and to the right of the equals sign you should enter the formula for f(t). Your answer must be mathematically valid. For example, if you believe that

f(t) = t^ – 3 + 5* t – 3,
then in the text box you should enter f(t) = t^(–3) + 5*t–3. After entering your answer, click the Save Answer button.

Response to your input section.
Enter The Value Of k Here:

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Evelyn 4 weeks 2021-12-26T10:58:37+00:00 1 Answer 0 views 0

Answers ( )

    0
    2021-12-26T11:00:09+00:00

    Answer:

    • f(t) = 6000*1.03^t
    • k = 0.0295588

    Step-by-step explanation:

    The formula for exponential growth can be written a couple of different ways. One I prefer uses the problem numbers directly:

      population = (initial population) × (growth factor)^t

    Here, the initial population is 6000, and the growth factor is 1+3% = 1.03. Then the function can be written as …

      f(t) = 6000·1.03^t

    __

    Another way to write the function is using the form …

      f(t) = (initial population) × e^(kt)

    where k is the natural logarithm of the growth factor. In this form, we have …

      k = ln(1.03)

      k ≈ 0.0295588

    so the function would be written …

      f(t) = 6000·e^(0.0295588t)

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