A national park begins tracking its wolf population in the year 2008. In 2008, t
ID: 3032621 • Letter: A
Question
A national park begins tracking its wolf population in the year 2008. In 2008, they found there were 200 wolves, and estimated that the population was growing exponentially by 3.5% per year. Let f be the function that assigns, to each time t after the year 2008, the total number of wolves in this national park. Find a formula for f. Find how many wolves are estimated to be in the park by 2018. Suppose for an exponential function h(x) = AB^x, you know that h(3) = 32 and h(5) = 128. Find the growth factor of h. (In other words, find B.) Show one step of work.Explanation / Answer
Dear Student Thank you for using Chegg Given Population of wolved rise exponentially after 2008 Let the General function be P = p0e^(kt) p :- Population in any year after 2008 P0 :- Population in 2008 (Given 200) t :- No of years after 2008 population is sought r :- Rate of rise of population (Given 3.5%) a) In 2009 P = P0 + 3.5% of P0 = 207 Wolves Calculating k 207 = 200*exp (kt) 1.035 = exp(k) 0.034401 = k (After taking Natural Logarithm) k = 0.034401 Hence general function is P = P0e^(0.034401t) b) Population in 2018 t = 10 years P = 282 wolves
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