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. From Year 1 to Year 4, the average rate of change of the population was about 62.4 bass per year.

From Year 5 to Year 8, the average rate of change of the population was about 67.6 bass per year.

The rate of change increased by about bass per a year.

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.

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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.

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6,250

step-by-step explanation:

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to find out how many bags of fertilizer it will take, first figure out how many square feet one bag can hold. divide 1000 by 8(125) to get the unit rate(how many square feet one pound of fertilizer can hold)and multiply it by 50(6250). if i haven't made a mistake, that is the answer.