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If [tex]$f(4)=246.4$[/tex] when [tex]$r=0.04$[/tex] for the function [tex]$f(t)=P e^{rt}$[/tex], then what is the approximate value of [tex]$P$[/tex]?

A. 289
B. 50
C. 210
D. 1220

Answer :

To solve this problem, we need to find the approximate value of [tex]\( P \)[/tex] for the function [tex]\( f(t) = P e^{rt} \)[/tex], where we know that [tex]\( f(4) = 246.4 \)[/tex] and [tex]\( r = 0.04 \)[/tex].

1. Set up the equation:
We have the function [tex]\( f(t) = P e^{rt} \)[/tex]. Given that [tex]\( f(4) = 246.4 \)[/tex], we substitute into the equation:
[tex]\[
246.4 = P e^{0.04 \times 4}
\][/tex]

2. Calculate the exponent:
You need to calculate [tex]\( 0.04 \times 4 \)[/tex], which equals 0.16. So, the equation becomes:
[tex]\[
246.4 = P e^{0.16}
\][/tex]

3. Solve for [tex]\( P \)[/tex]:
To find [tex]\( P \)[/tex], divide both sides of the equation by [tex]\( e^{0.16} \)[/tex]:
[tex]\[
P = \frac{246.4}{e^{0.16}}
\][/tex]

4. Calculate [tex]\( e^{0.16} \)[/tex]:
The value of [tex]\( e^{0.16} \)[/tex] is approximately 1.1735.

5. Final calculation for [tex]\( P \)[/tex]:
Now we can find [tex]\( P \)[/tex]:
[tex]\[
P = \frac{246.4}{1.1735} \approx 209.97
\][/tex]

So, the approximate value of [tex]\( P \)[/tex] is 210. Therefore, the correct answer is option C: 210.

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