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The function [tex]$P(t)=3 \cdot e^{0.0866 t}$[/tex] describes how to find the pollution level [tex]$P(t)$[/tex] in relation to any time [tex][tex]$t$[/tex][/tex], where [tex]$t$[/tex] is the number of years after 1970.

According to this information, what was the pollution level in 1980? (Round your answer to the nearest hundredth.)

A. 2.38
B. 7.09
C. 2.36
D. 7.13


Sagot :

To determine the pollution level in 1980 using the function [tex]\( P(t) = 3 \cdot e^{0.0866 t} \)[/tex], we need to follow a series of steps:

1. Identify the value of [tex]\( t \)[/tex] for the year 1980. The variable [tex]\( t \)[/tex] represents the number of years after 1970.
[tex]\[ t = 1980 - 1970 = 10. \][/tex]

2. Substitute [tex]\( t = 10 \)[/tex] into the function [tex]\( P(t) = 3 \cdot e^{0.0866 t} \)[/tex]:
[tex]\[ P(10) = 3 \cdot e^{0.0866 \cdot 10}. \][/tex]

3. Calculate the exponent:
[tex]\[ 0.0866 \times 10 = 0.866. \][/tex]

4. Find the value of the exponent using [tex]\( e \)[/tex] (Euler's number, approximately 2.71828):
[tex]\[ P(10) = 3 \cdot e^{0.866}. \][/tex]

5. Compute [tex]\( e^{0.866} \)[/tex]:
[tex]\[ e^{0.866} \approx 2.377382279962065. \][/tex]

6. Multiply this result by 3:
[tex]\[ P(10) = 3 \cdot 2.377382279962065 \approx 7.132146839886197. \][/tex]

7. Round the result to the nearest hundredth:
[tex]\[ 7.13. \][/tex]

Therefore, the pollution level in 1980, rounded to the nearest hundredth, is [tex]\( 7.13 \)[/tex].

So the correct answer is:
D. 7.13
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