Question

one model of earths population growth is $p(t)=\frac{64}{(1 + 11e^{-0.08t})}$, where t is measured in years since 1990, and p is measured in billions of people. which of the following statements are true? check all that apply. the population of earth will grow exponentially for a while but then start to slow down its growth. the carrying capacity of earth is 5.33 billion people. the population of earth is increasing by a steady rate of 8% per year. in 1990, there were 5.33 billion people.

Explanation:

Step1: Analyze the logistic function form

The given function \(P(t)=\frac{64}{1 + 11e^{-0.08t}}\) is a logistic function of the form \(P(t)=\frac{K}{1+Ae^{-rt}}\), where \(K\) is the carrying capacity, \(A\) is a constant, and \(r\) is the growth rate. Logistic functions show exponential - like growth initially (when \(t\) is small, \(e^{-rt}\approx1 - rt\) for small \(rt\)) and then the growth slows down as \(t\) increases (since \(e^{-rt}\to0\) as \(t\to\infty\) and \(P(t)\to K\)).

Step2: Find the carrying capacity

For the logistic function \(P(t)=\frac{K}{1+Ae^{-rt}}\), comparing with \(P(t)=\frac{64}{1 + 11e^{-0.08t}}\), we have \(K = 64\) (carrying capacity).

Step3: Check the growth rate interpretation

The formula for the population growth in a logistic model is not a simple \(P(t)=P_0(1 + r)^t\) (exponential growth formula). The \(r = 0.08\) in the logistic function \(P(t)=\frac{64}{1 + 11e^{-0.08t}}\) is not the same as a simple annual percentage increase in an exponential - like \(y = a(1 + p)^t\) formula.

Step4: Find the population in 1990

When \(t = 0\) (year 1990), \(P(0)=\frac{64}{1+11e^{0}}=\frac{64}{1 + 11}=\frac{64}{12}\approx5.33\).

Answer:

• The population of Earth will grow exponentially for a while but then start to slow down its growth.
• In 1990, there were 5.33 billion people.