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A pollutant in a river has an initial concentration of 3 parts per million and degrades at a rate of 1.5% per year. Approximate its concentration after 20 years.

User Canoe
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Final answer:

To find the concentration of a pollutant in a river after 20 years with an initial concentration of 3 ppm degrading at a rate of 1.5% per year, we use the exponential decay formula, which results in an approximate concentration of 2.22 ppm.

Step-by-step explanation:

To calculate the approximate concentration of a pollutant after a certain amount of time, we can use the formula which accounts for exponential decay: C(t) = C0 * e^(-kt), where C(t) is the concentration at time t, C0 is the initial concentration, e is the base of the natural logarithm, and k is the decay constant. Since we have a decay rate of 1.5% per year, we can convert this percentage into a decimal for k by dividing by 100, which gives us k = 0.015. Now, we can calculate the concentration after 20 years:

C(20) = 3 * e^(-0.015*20)

To solve for C(20), we need to calculate the expression e^(-0.015*20), which is approximately e^(-0.3). After computing this value, we multiply it by the initial concentration 3 ppm to get the concentration after 20 years:

C(20) ≈ 3 * e^(-0.3)

Using a calculator, e^(-0.3) ≈ 0.7408, and thus:

C(20) ≈ 3 * 0.7408 ≈ 2.2224 ppm

The approximate concentration of the pollutant in the river after 20 years is about 2.22 ppm.

User Dace Zarina
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