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Answer :
We are given the function
[tex]$$
f(t) = P e^{rt},
$$[/tex]
with the value of the rate as [tex]$r = 0.03$[/tex]. At time [tex]$t = 3$[/tex], the function takes the value
[tex]$$
f(3) = 191.5.
$$[/tex]
Substitute [tex]$t = 3$[/tex] and [tex]$r = 0.03$[/tex] into the function:
[tex]$$
f(3) = P e^{0.03 \times 3} = P e^{0.09}.
$$[/tex]
We know that [tex]$f(3) = 191.5$[/tex], so we have:
[tex]$$
191.5 = P e^{0.09}.
$$[/tex]
To solve for [tex]$P$[/tex], isolate it by dividing both sides by [tex]$e^{0.09}$[/tex]:
[tex]$$
P = \frac{191.5}{e^{0.09}}.
$$[/tex]
Evaluating the denominator, we find that:
[tex]$$
e^{0.09} \approx 1.09417.
$$[/tex]
Thus,
[tex]$$
P \approx \frac{191.5}{1.09417} \approx 175.
$$[/tex]
Therefore, the approximate value of [tex]$P$[/tex] is [tex]$\boxed{175}$[/tex].
[tex]$$
f(t) = P e^{rt},
$$[/tex]
with the value of the rate as [tex]$r = 0.03$[/tex]. At time [tex]$t = 3$[/tex], the function takes the value
[tex]$$
f(3) = 191.5.
$$[/tex]
Substitute [tex]$t = 3$[/tex] and [tex]$r = 0.03$[/tex] into the function:
[tex]$$
f(3) = P e^{0.03 \times 3} = P e^{0.09}.
$$[/tex]
We know that [tex]$f(3) = 191.5$[/tex], so we have:
[tex]$$
191.5 = P e^{0.09}.
$$[/tex]
To solve for [tex]$P$[/tex], isolate it by dividing both sides by [tex]$e^{0.09}$[/tex]:
[tex]$$
P = \frac{191.5}{e^{0.09}}.
$$[/tex]
Evaluating the denominator, we find that:
[tex]$$
e^{0.09} \approx 1.09417.
$$[/tex]
Thus,
[tex]$$
P \approx \frac{191.5}{1.09417} \approx 175.
$$[/tex]
Therefore, the approximate value of [tex]$P$[/tex] is [tex]$\boxed{175}$[/tex].
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