Question 296092
A new farm pond was stocked with 2500 crappies in 2003. 
(0,2500)
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The crappie population in 2006 was found to be 4320.
(3,4320)
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a) Let t be the number of years after 2003 (in other words t = 0 corresponds to 2003). Write down the initial crappie population at t = 3.
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b) Find the growth function of the form f(t) = yo*b^t that gives the crappie population t years after 2003. The function you have found models 
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2500 = yo
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4320 = 2500*b^3
1.729 = b^3
b = 1.2
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Model: f(t) = 2500*(1.2)^t
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c) Predict the crappie population in 2010.
f(10) = 2500*1.2^10
f(10) = 15479
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d) In what year will the crappie population reach 13,000? Give the exact value for t and then use your calculator to approximate t to get the year.
13000 = 2500*1.2^t
5.2 = 1.2^t
t*ln(1.2) = ln(5.2)
t = [ln(5.2)/ln(1.2)]
t = 17.2995...
t is appox 17 years
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e) Are your answers for (c) and (d) consistent with the given population data in your answer to (b)?
Comment: I'll leave that opinion to you.
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Cheers,
Stan H.