Final answer:
To predict the population of Cary in 2008 using the uninhibited growth model, first calculate the growth rate based on 1980 and 1987 data, and then apply the rate to predict the 2008 population. The predicted population is approximately 119,700.
Step-by-step explanation:
The student asks for a prediction of the population of Cary for the year 2008 using the uninhibited growth model. Here, we'll use an exponential growth model since the assumption is that the population grows without bounds. The general formula for exponential growth is P(t) = P0ert, where P(t) is the population at time t, P0 is the initial population, r is the growth rate, and t is the time in years since the initial population was measured.
To predict the population for Cary in the year 2008, we'll first need to find the growth rate (r). We do this by using the provided population data for Cary from 1980 and 1987. We will then apply the growth rate to predict the population for 2008.
First, we determine 'r' using the formula:
39387 = 21763er(7), where 7 is the number of years between 1980 and 1987.
Next, we solve for r:
r = ln(39387/21763)/7 ≈ 0.0609
Lastly, we use the growth rate to predict the population for 2008:
P(2008) = 21763e0.0609(28) (since 2008 is 28 years from 1980).
This results in:
P(2008) = 21763e1.7052 ≈ 21763 * 5.503 ≈ 119,700 (rounded to the nearest whole number).
Therefore, the predicted population of Cary in 2008 using the uninhibited growth model is approximately 119,700.