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A small country emits 54,000 kilotons of carbon dioxide per year. In a recent global agreement, the country agreed to cut its carbon emissions by 
4% per year for the next 14 years. In the first year of the agreement, the country will keep its emissions at 54,000 kilotons and the emissions will decrease 
4% in each successive year. How many total kilotons of carbon dioxide would the country emit over the course of the 14 year period, to the nearest whole number?
Answer:

A small country emits 5454,000000 kilotons of carbon dioxide per year. In a recent global agreement, the country agreed to cut its carbon emissions by 4% 4 \% per year for the next 1414 years. In the first year of the agreement, the country will keep its emissions at 5454,000000 kilotons and the emissions will decrease 4% 4 \% in each successive year. How many total kilotons of carbon dioxide would the country emit over the course of the 1414 year period, to the nearest whole number?\newlineAnswer:

Full solution

Q. A small country emits 5454,000000 kilotons of carbon dioxide per year. In a recent global agreement, the country agreed to cut its carbon emissions by 4% 4 \% per year for the next 1414 years. In the first year of the agreement, the country will keep its emissions at 5454,000000 kilotons and the emissions will decrease 4% 4 \% in each successive year. How many total kilotons of carbon dioxide would the country emit over the course of the 1414 year period, to the nearest whole number?\newlineAnswer:
  1. Understand the problem: Understand the problem.\newlineWe need to calculate the total carbon emissions over 1414 years, with a 4%4\% reduction each year after the first year. The initial emission is 54,00054,000 kilotons.
  2. Calculate emissions for each year: Calculate the emissions for each year.\newlineThe emissions for the first year are 54,00054,000 kilotons. For each subsequent year, the emissions are 4%4\% less than the previous year.
  3. Use sum formula for series: Use the formula for the sum of a decreasing geometric series.\newlineThe sum SS of nn terms of a geometric series with first term aa and common ratio rr is given by S=a(1rn)(1r)S = \frac{a(1 - r^n)}{(1 - r)}, where 0<r<10 < r < 1.\newlineIn this case, a=54,000a = 54,000 kilotons, r=0.96r = 0.96 (since there's a 44% reduction, 100%4%=96%=0.96100\% - 4\% = 96\% = 0.96), and n=14n = 14.
  4. Calculate total emissions: Calculate the total emissions.\newlineS=54,000×(10.9614)/(10.96)S = 54,000 \times (1 - 0.96^{14}) / (1 - 0.96)\newlineFirst, calculate 0.96140.96^{14}.
  5. Calculate 0.96140.96^{14}: Calculate 0.96140.96^{14}. \newline0.96140.5630.96^{14} \approx 0.563
  6. Substitute into sum formula: Substitute 0.96140.96^{14} into the sum formula.\newlineS=54,000×(10.563)/(10.96)S = 54,000 \times (1 - 0.563) / (1 - 0.96)
  7. Calculate numerator of sum: Calculate the numerator of the sum formula.\newline10.563=0.4371 - 0.563 = 0.437\newlineS=54,000×0.437/(10.96)S = 54,000 \times 0.437 / (1 - 0.96)
  8. Calculate denominator of sum: Calculate the denominator of the sum formula.\newline10.96=0.041 - 0.96 = 0.04\newlineS=54,000×0.437/0.04S = 54,000 \times 0.437 / 0.04
  9. Calculate total emissions: Calculate the total emissions SS. \newlineS=54,000×0.437/0.04S = 54,000 \times 0.437 / 0.04\newlineS=54,000×10.925S = 54,000 \times 10.925\newlineS590,550 kilotonsS \approx 590,550 \text{ kilotons}
  10. Round total emissions: Round the total emissions to the nearest whole number. S590,550S \approx 590,550 kilotons

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