Answer :
To solve the problem, we begin with the function [tex]\( f(t) = \rho e^{rt} \)[/tex]. We are given that [tex]\( f(4) = 246.4 \)[/tex] when [tex]\( r = 0.04 \)[/tex].
1. Set up the equation with the provided values:
The equation becomes [tex]\( \rho e^{0.04 \times 4} = 246.4 \)[/tex].
2. Calculate the exponent:
First, we calculate [tex]\( 0.04 \times 4 = 0.16 \)[/tex].
3. Compute [tex]\( e^{0.16} \)[/tex]:
The value of [tex]\( e^{0.16} \)[/tex] is approximately 1.1735.
4. Solve for [tex]\( \rho \)[/tex]:
We rearrange the equation to find [tex]\( \rho \)[/tex]:
[tex]\[
\rho = \frac{246.4}{e^{0.16}}
\][/tex]
[tex]\[
\rho = \frac{246.4}{1.1735} \approx 209.97
\][/tex]
5. Choose the closest answer:
The value of [tex]\( \rho \)[/tex] is approximately 209.97, which is closest to 210. Therefore, the approximate value of [tex]\( P \)[/tex] is:
A. 210
1. Set up the equation with the provided values:
The equation becomes [tex]\( \rho e^{0.04 \times 4} = 246.4 \)[/tex].
2. Calculate the exponent:
First, we calculate [tex]\( 0.04 \times 4 = 0.16 \)[/tex].
3. Compute [tex]\( e^{0.16} \)[/tex]:
The value of [tex]\( e^{0.16} \)[/tex] is approximately 1.1735.
4. Solve for [tex]\( \rho \)[/tex]:
We rearrange the equation to find [tex]\( \rho \)[/tex]:
[tex]\[
\rho = \frac{246.4}{e^{0.16}}
\][/tex]
[tex]\[
\rho = \frac{246.4}{1.1735} \approx 209.97
\][/tex]
5. Choose the closest answer:
The value of [tex]\( \rho \)[/tex] is approximately 209.97, which is closest to 210. Therefore, the approximate value of [tex]\( P \)[/tex] is:
A. 210
