Points] DETAILS MY NOTES SPRECALC8 4.6.007. This exercise uses the exponential growth model. The population of a country has a relative growth rate of \( 2 \% \) per year. The government is trying to reduce the growth rate to \( 1 \% \). The population in 2011 was approximately 120 million. Find the projected population (in million people) for the year 2033 for the following conditions. (Round your answers to the nearest million.) (a) The relative growth rate remains at \( 2 \% \) per year. million people (b) The relative growth rate is reduced to \( 1 \% \) per year. million people

To project the population using the exponential growth model, we utilize the formula: \[ P(t) = P_0 \times e^{rt} \] where \( P(t) \) is the population at time \( t \), \( P_0 \) is the initial population, \( r \) is the relative growth rate, and \( t \) is the time in years. For part (a), with a growth rate of \( 2\% \): - Initial population, \( P_0 = 120 \) million - \( r = 0.02 \) - Time period from 2011 to 2033 is \( t = 22 \) years. Calculating: \[ P(22) = 120 \times e^{0.02 \times 22} \approx 120 \times e^{0.44} \approx 120 \times 1.552 \approx 186.24 \] Rounding gives approximately **186 million people**. For part (b), with a reduced growth rate of \( 1\% \): - Initial population, \( P_0 = 120 \) million - \( r = 0.01 \) Calculating: \[ P(22) = 120 \times e^{0.01 \times 22} \approx 120 \times e^{0.22} \approx 120 \times 1.246 \approx 149.52 \] Rounding gives approximately **150 million people**.