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A small country emits 80,000 kilotons of carbon dioxide per year. In a recent global
agreement, the country agreed to cut its carbon emissions by 2.6% per year for the
next 11 years. In the first year of the agreement, the country will keep its emissions at
80,000 kilotons and the emissions will decrease 2.6% in each successive year. How
many total kilotons of carbon dioxide would the country emit over the course of the
11 year period, to the nearest whole number?

Sagot :

Using an exponential function and integration, it is found that over the entire 11 year period, the country will emit 763,968 kilotons of carbon dioxide.

Exponential function:

A decaying exponential function is modeled by:

[tex]A(t) = A(0)(1 - r)^t[/tex]

In which:

  • A(0) is the initial value.
  • r is the decay rate, as a decimal.

In this problem:

  • In the first year, the country emits 80,000 kilotons, hence [tex]A(0) = 80000[/tex].
  • Emissions will decrease 2.6% in each successive year, hence [tex]r = 0.026[/tex]

Then, the equation for the amount emitted each year is:

[tex]A(t) = A(0)(1 - r)^t[/tex]

[tex]A(t) = 80000(1 - 0.026)^t[/tex]

[tex]A(t) = 80000(0.974)^t[/tex]

The total amount emitted over the course of the 11 year period is:

[tex]T = \int_{0}^{11} A(t) dt[/tex]

Hence:

[tex]T = \int_{0}^{11} 80000(0.974)^t dt[/tex]

[tex]T = \frac{80000(0.974)^t}{\ln{0.974}}|_{t = 0}^{t = 11}[/tex]

Applying the Fundamental Theorem of Calculus:

[tex]T = \frac{80000(0.974)^{11}}{\ln{0.974}} - \frac{80000(0.974)^0}{\ln{0.974}} = \frac{80000}{\ln{0.974}}((0.974)^{11} - 1) = 763968[/tex]

Over the entire 11 year period, the country will emit 763,968 kilotons of carbon dioxide.

You can learn more about exponential functions at https://brainly.com/question/25537936

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