a. in 2000, the population of a country was approximately 5.83 million and by 2060 it is projected to grow…

a. in 2000, the population of a country was approximately 5.83 million and by 2060 it is projected to grow to 11 million. use the exponential growth model ( a = a_0e^{kt} ), in which ( t ) is the number of years after 2000 and ( a_0 ) is in millions, to find an exponential growth function that models the data.\nb. by which year will the population be 8 million?\na. the exponential growth function that models the data is ( a=square )\n(simplify your answer. use integers or decimals for any numbers in the expression. round to two decimal places as needed.)

a. in 2000, the population of a country was approximately 5.83 million and by 2060 it is projected to grow to 11 million. use the exponential growth model ( a = a_0e^{kt} ), in which ( t ) is the number of years after 2000 and ( a_0 ) is in millions, to find an exponential growth function that models the data.\nb. by which year will the population be 8 million?\na. the exponential growth function that models the data is ( a=square )\n(simplify your answer. use integers or decimals for any numbers in the expression. round to two decimal places as needed.)

Answer

Explanation:

Step1: Find the value of (k)

Given (A_0 = 5.83) (population in 2000, (t = 0)), and in 2060 ((t=60)), (A = 11). Substitute into the formula (A = A_0e^{kt}): (11=5.83e^{k\times60}) (\frac{11}{5.83}=e^{60k}) Take the natural logarithm of both sides: (\ln(\frac{11}{5.83})=\ln(e^{60k})) Since (\ln(e^{x}) = x), we have (60k=\ln(\frac{11}{5.83})) (k=\frac{\ln(\frac{11}{5.83})}{60}) Calculate (\ln(\frac{11}{5.83})\approx\ln(1.8868)= 0.635) (k=\frac{0.635}{60}\approx0.01)

Step2: Write the exponential growth function

Substitute (A_0 = 5.83) and (k\approx0.01) into (A = A_0e^{kt}) (A = 5.83e^{0.01t})

Answer:

(A = 5.83e^{0.01t})