Problem

National Park Service personnel are trying to increase the size of the bison population of the national park. If 211 bison currently live in the park, and if the population's rate of growth is $4 \%$ annually, find how many bison there should be in 8 years.
There should be approximately bison in 8 years. (Round to the nearest whole number as needed.)

Answer

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Answer

Final Answer: There should be approximately \(\boxed{289}\) bison in 8 years.

Steps

Step 1 :The problem is asking for the future value of the bison population given an initial population, a growth rate, and a time period. This is a compound interest problem, where the interest is compounded annually. The formula for compound interest is: \(A = P(1 + \frac{r}{n})^{nt}\) where: \(A\) is the amount of money earned after n years, including interest, \(P\) is the principal amount (the initial amount of money), \(r\) is the annual interest rate (in decimal form), \(n\) is the number of times that interest is compounded per year, and \(t\) is the time the money is invested for, in years.

Step 2 :In this case, \(P\) is the initial bison population (211), \(r\) is the annual growth rate (4% or 0.04), \(n\) is 1 (since the population grows annually), and \(t\) is 8 years. We need to find \(A\), the future value of the bison population.

Step 3 :Substitute the given values into the formula: \(A = 211(1 + \frac{0.04}{1})^{1*8}\)

Step 4 :Solve the equation to find the future value of the bison population: \(A = 289\)

Step 5 :Final Answer: There should be approximately \(\boxed{289}\) bison in 8 years.

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