Question

Compound Interest An investment of $7000 is deposited into an account in which interest is compounded continuously. Complete the table by filling in the amounts to which the investment grows at the indicated times or interest rates.

$r=3\%$

Short answer

Time (years)	Amount ($)
1	7213.18
2	7432.86
3	7659.22
4	7892.48
5	8132.84
6	8380.52

Step-by-step solution

Step 1. Given.

$P=\mathrm{Principal}=7000$.

$r=\mathrm{interest}\mathrm{rate}=3\%=0.03$.

Step 2. To determine.

We have to complete the table by filling in the amounts to which the investment grows at the indicated times.

Step 3. Calculation.

Given:

$P=\mathrm{Principal}=7000$.

$r=\mathrm{interest}\mathrm{rate}=4\%=0.04$.

$t=\mathrm{number}\mathrm{of}\mathrm{years}$.

$A\left(t\right)=\mathrm{Amount}\mathrm{after}t\mathrm{years}$

Here,

$$\begin{gathered} A\left(t\right)=P{e}^{rt} \\ =7000e^{0.03t} \end{gathered}$$

We have to find $A\left(t\right)$for $t=1,2,3,4,5,6$. We plug these values in $A\left(t\right)$.

$$\begin{gathered} A\left(1\right)=7000e^{0.03\left(1\right)} \\ =7213.18173767 \\ \approx 7213.18 \end{gathered}$$

$$\begin{gathered} A\left(2\right)=7000e^{0.03\left(2\right)} \\ =7432.85582582 \\ \approx 7432.86 \end{gathered}$$

$$\begin{gathered} A\left(3\right)=7000e^{0.03\left(3\right)} \\ =7659.21998594 \\ \approx 7659.22 \end{gathered}$$

$$\begin{gathered} A\left(4\right)=7000e^{0.03\left(4\right)} \\ =7892.47796106 \\ \approx 7892.48 \end{gathered}$$

$$\begin{gathered} A\left(5\right)=7000e^{0.03\left(5\right)} \\ =8132.8396991 \\ \approx 8132.84 \end{gathered}$$

$$\begin{gathered} A\left(6\right)=7000e^{0.03\left(6\right)} \\ =8380.52154185 \\ \approx 8380.52 \end{gathered}$$

Hence the table is:

Time (years)	Amount ($)
1	7213.18
2	7432.86
3	7659.22
4	7892.48
5	8132.84
6	8380.52