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
