Answer :
- Substitute the given values into the function: $191.5 = P e^{0.03 \times 3}$.
- Solve for $P$: $P = \frac{191.5}{e^{0.09}}$.
- Approximate the value of $e^{0.09}$: $e^{0.09} \approx 1.09417$.
- Calculate $P$: $P \approx \frac{191.5}{1.09417} \approx 175$. The final answer is $\boxed{175}$.
### Explanation
1. Problem Setup
We are given the function $f(t) = P e^t$ and the information that $f(3) = 191.5$. Our goal is to find the approximate value of $P$.
2. Substitution
We substitute the given values into the function: $191.5 = P e^3$.
3. Isolating P
Now, we solve for $P$ by dividing both sides of the equation by $e^3$: $$P = \frac{191.5}{e^3}$$.
4. Approximation
We know that $e^3 \approx 20.0855$. Therefore, $$P \approx \frac{191.5}{20.0855} \approx 9.5342$$.
5. Corrected Approximation
Since we are looking for the approximate value of $P$ from the given options, we can see that the closest value to $9.5342$ is not among the options. However, there seems to be an error in the problem statement. It says $r=0.03$, but $r$ is not used in the function $f(t) = P e^t$. It should be $t=3$. Let's assume that the function is $f(t) = P e^{rt}$ and $t=3$ and $r=0.03$. Then $f(3) = P e^{0.03 Imes 3} = P e^{0.09} = 191.5$. So $P = \frac{191.5}{e^{0.09}}$. We know that $e^{0.09} \approx 1.09417$. Therefore, $P \approx \frac{191.5}{1.09417} \approx 175$.
6. Final Answer
The approximate value of $P$ is 175.
### Examples
Exponential functions are used to model population growth, radioactive decay, and compound interest. For example, if you invest money in an account that earns compound interest, the amount of money you have after a certain amount of time can be modeled by an exponential function. Understanding how to solve for the initial investment (P) given the final amount and interest rate is crucial in financial planning.
- Solve for $P$: $P = \frac{191.5}{e^{0.09}}$.
- Approximate the value of $e^{0.09}$: $e^{0.09} \approx 1.09417$.
- Calculate $P$: $P \approx \frac{191.5}{1.09417} \approx 175$. The final answer is $\boxed{175}$.
### Explanation
1. Problem Setup
We are given the function $f(t) = P e^t$ and the information that $f(3) = 191.5$. Our goal is to find the approximate value of $P$.
2. Substitution
We substitute the given values into the function: $191.5 = P e^3$.
3. Isolating P
Now, we solve for $P$ by dividing both sides of the equation by $e^3$: $$P = \frac{191.5}{e^3}$$.
4. Approximation
We know that $e^3 \approx 20.0855$. Therefore, $$P \approx \frac{191.5}{20.0855} \approx 9.5342$$.
5. Corrected Approximation
Since we are looking for the approximate value of $P$ from the given options, we can see that the closest value to $9.5342$ is not among the options. However, there seems to be an error in the problem statement. It says $r=0.03$, but $r$ is not used in the function $f(t) = P e^t$. It should be $t=3$. Let's assume that the function is $f(t) = P e^{rt}$ and $t=3$ and $r=0.03$. Then $f(3) = P e^{0.03 Imes 3} = P e^{0.09} = 191.5$. So $P = \frac{191.5}{e^{0.09}}$. We know that $e^{0.09} \approx 1.09417$. Therefore, $P \approx \frac{191.5}{1.09417} \approx 175$.
6. Final Answer
The approximate value of $P$ is 175.
### Examples
Exponential functions are used to model population growth, radioactive decay, and compound interest. For example, if you invest money in an account that earns compound interest, the amount of money you have after a certain amount of time can be modeled by an exponential function. Understanding how to solve for the initial investment (P) given the final amount and interest rate is crucial in financial planning.
