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A major fishing company does its fishing in a local lake. The first year of the company's operations it managed to catch 70,000 fish. Due to population decreases, the number of fish the company was able to catch decreased by 3% each year. How many total fish did the company catch over the first 10 years, to the nearest whole number? Answer:

A major fishing company does its fishing in a local lake. The first year of the company's operations it managed to catch 7070,000000 fish. Due to population decreases, the number of fish the company was able to catch decreased by 3% 3 \% each year. How many total fish did the company catch over the first 1010 years, to the nearest whole number?\newlineAnswer:

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Q. A major fishing company does its fishing in a local lake. The first year of the company's operations it managed to catch 7070,000000 fish. Due to population decreases, the number of fish the company was able to catch decreased by 3% 3 \% each year. How many total fish did the company catch over the first 1010 years, to the nearest whole number?\newlineAnswer:
  1. Identify data: Identify the initial amount of fish caught and the annual decrease rate.\newlineThe initial amount of fish caught is 70,00070,000, and the annual decrease rate is 3%3\%.
  2. Calculate using formula: Calculate the number of fish caught each year using the formula for exponential decay.\newlineThe formula for exponential decay is P(t)=P0×(1−r)tP(t) = P_0 \times (1 - r)^t, where P(t)P(t) is the population at time tt, P0P_0 is the initial population, rr is the decay rate, and tt is the time in years.
  3. Calculate for 1010 years: Calculate the number of fish caught each year for 1010 years.\newlineYear 11: 70,000×(1−0.03)170,000 \times (1 - 0.03)^1\newlineYear 22: 70,000×(1−0.03)270,000 \times (1 - 0.03)^2\newline...\newlineYear 1010: 70,000×(1−0.03)1070,000 \times (1 - 0.03)^{10}
  4. Perform calculations: Perform the calculations for each year.\newlineYear 11: 70,000×0.97=67,90070,000 \times 0.97 = 67,900\newlineYear 22: 70,000×0.972≈65,88370,000 \times 0.97^2 \approx 65,883\newline...\newlineYear 1010: 70,000×0.9710≈49,02570,000 \times 0.97^{10} \approx 49,025
  5. Sum total fish caught: Sum the total number of fish caught over the 1010 years.\newlineTotal fish caught =Year 1+Year 2+…+Year 10= \text{Year } 1 + \text{Year } 2 + \ldots + \text{Year } 10
  6. Perform summation calculation: Perform the summation calculation.\newlineTotal fish caught ≈67,900+65,883+…+49,025\approx 67,900 + 65,883 + \ldots + 49,025
  7. Use calculator for sum: Use a calculator or software to find the sum of the geometric series.\newlineTotal fish caught ≈67,900+65,883+64,907+62,970+61,072+59,210+57,384+55,593+53,835+52,111≈600,865\approx 67,900 + 65,883 + 64,907 + 62,970 + 61,072 + 59,210 + 57,384 + 55,593 + 53,835 + 52,111 \approx 600,865

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