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If [tex]f(3)=191.5[/tex] when [tex]r=0.03[/tex] for the function [tex]f(t)=P e^f[/tex], then what is the approximate value of [tex]P[/tex]?

A. 175
B. 210
C. 471
D. 78

Answer :

We are given the function
[tex]$$
f(t) = P e^{rt}
$$[/tex]
with [tex]$r = 0.03$[/tex]. Since [tex]$f(3) = 191.5$[/tex], we substitute into the equation:
[tex]$$
191.5 = P e^{0.03 \times 3}.
$$[/tex]

First, compute the exponent:
[tex]$$
0.03 \times 3 = 0.09.
$$[/tex]

Now the equation becomes:
[tex]$$
191.5 = P e^{0.09}.
$$[/tex]

To solve for [tex]$P$[/tex], divide both sides by [tex]$e^{0.09}$[/tex]:
[tex]$$
P = \frac{191.5}{e^{0.09}}.
$$[/tex]

Using the approximate value:
[tex]$$
e^{0.09} \approx 1.0941742837052104,
$$[/tex]
we substitute to get:
[tex]$$
P \approx \frac{191.5}{1.0941742837052104} \approx 175.01782197944019.
$$[/tex]

Thus, the value of [tex]$P$[/tex] is approximately [tex]$175$[/tex], which corresponds to option A.

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