Question

There are approximately 3,000 bass in a lake. The population grows at a rate of 2% per year. Round answers to the nearest tenth.

Answer 1

4.8 is the answer (sorry if this doesn’t work)

Answer 2

Answer:

years 1–4: 62.4 bass per year

years 5–8: 67.6 bass per year

Step-by-step explanation:

If the population in year n is ...

 p(n) = 3000·1.02^n

then the average rate of change from year 1 to year 4 is ...

 (p(4) -p(1))/(4 -1) = 3000(1.02^4 -1.02^1)/3 = 1020·(1.02^3 -1) ≈ 62.4

The average rate of change for years 1–4 is 62.4 bass per year.

For years 5–8, the rate of change is similarly computed:

 (p(8) -p(5))/(8 -5) = 3000(1.02^8 -1.02^5)/3 = 1000·1.02^5·(1.02^3 -1) ≈ 67.6

The average rate of change for years 5–8 is 67.6 bass per year.