Answer:
The best prediction for the number of years it will take for the population to reach 200,000 is 9.41
Explanation:
Year Population
1 11,920
2 16,800
3 23,300
4 33,000
5 45,750
6 64,000

Substitute the values

Divide 1 and 2

The exponential function that passes through the points (1, 11920) and (2, 16800) is

Now we are supposed to find the best prediction for the number of years it will take for the population to reach 200,000



t = 9.41
Hence the best prediction for the number of years it will take for the population to reach 200,000 is 9.41