Question

how much would you need to deposit in an account now in order to have $3000 in the account in 15 years? assume the account earns 3% interest compounded monthly. round to the nearest penny if needed.
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Explanation:

Step1: Recall the compound - interest formula for present value

The formula for compound interest is $A = P(1+\frac{r}{n})^{nt}$, where:

• $A$ is the future value of the investment/loan, including interest. Here, $A=\$3000$.
• $P$ is the principal investment amount (the initial deposit or loan amount), which we need to find.
• $r$ is the annual interest rate (decimal). Here, $r = 0.03$ (since $3\%=0.03$).
• $n$ is the number of times that interest is compounded per year. Since it is compounded monthly, $n = 12$.
• $t$ is the time the money is invested or borrowed for, in years. Here, $t = 15$ years.

We need to solve for $P$, so we can re - arrange the formula to $P=\frac{A}{(1 + \frac{r}{n})^{nt}}$.

Step2: Substitute the given values into the formula

Substitute $A = 3000$, $r=0.03$, $n = 12$, and $t = 15$ into the formula for $P$:

First, calculate the exponent $nt$: $nt=12\times15 = 180$.

Then, calculate the value of $(1+\frac{r}{n})$: $\frac{r}{n}=\frac{0.03}{12}=0.0025$, and $1+\frac{r}{n}=1 + 0.0025=1.0025$.

Next, calculate $(1+\frac{r}{n})^{nt}=(1.0025)^{180}$.

We can calculate $(1.0025)^{180}$ using a calculator. $(1.0025)^{180}\approx1.55303071$

Then, $P=\frac{3000}{1.55303071}\approx1931.07$

Answer:

\$1931.07