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Ostrovityanka [42]
3 years ago
13

PLEASEEEEE HELPPPPP MEEEE!!!!

Mathematics
1 answer:
amid [387]3 years ago
8 0
The answer should be c
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Circumference / diameter = pi / 1

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4 0
4 years ago
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.
Tamiku [17]

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.

5 0
3 years ago
Read 2 more answers
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