Question

The population of a city has risen and fallen over a 20-year interval. Its population may be modeled by the function y = 15,000 + 4,000 sin(0.637x), where the domain is the years since 1980 and the range is the population of the city. Assuming this trend continues, what are the predicted populations in 2007 and 2010? (Round your answers to the nearest whole number.)

Answer 1

To find the predicted population in 2007 and 2010, substitute the given x-values in the function y = 15,000 + 4,000 sin(0.637x). The predicted population in 2007 is approximately 19,708, and in 2010 it is approximately 19,075.

Step-by-step explanation:

To find the predicted population in 2007 and 2010, we can substitute the given x-values in the function y = 15,000 + 4,000 sin(0.637x).

a. When x = 2007, the calculation would be: y = 15,000 + 4,000 sin(0.637 ✕ 27). Evaluating this expression gives us a population of approximately 19,708.

b. When x = 2010, the calculation would be: y = 15,000 + 4,000 sin(0.637 ✕ 30). Evaluating this expression gives us a population of approximately 19,075.

Answer 2

To find the predicted population in 2007 and 2010, we need to substitute the corresponding values of x into the given function y = 15,000 + 4,000 sin(0.637x). For x = 2007 - 1980 = 27, the predicted population in 2007 is 18,600. For x = 2010 - 1980 = 30, the predicted population in 2010 is 18,223.

Step-by-step explanation:

To find the predicted population in 2007 and 2010, we need to substitute the corresponding values of x into the given function y = 15,000 + 4,000 sin(0.637x).

For x = 2007 - 1980 = 27:

y = 15,000 + 4,000 sin(0.637 * 27) = 15,000 + 4,000 sin(17.199) = 15,000 + 3,600 = 18,600.

So, the predicted population in 2007 is 18,600.

For x = 2010 - 1980 = 30:

y = 15,000 + 4,000 sin(0.637 * 30) = 15,000 + 4,000 sin(19.11) = 15,000 + 3,223 = 18,223.

So, the predicted population in 2010 is 18,223.