To find the amount in the account at the end of one year, we need to use the compound interest formula $A = P\left(1+\frac{r}{n}\right)^{nt}$, where:
$A$ is the amount in the account at the end of one year,
$P$ is the principal amount (initial deposit) of $3,500,
$r$ is the annual interest rate as a decimal (3% or 0.03),
$n$ is the number of times that interest is compounded per year (in this case once per year), and
$t$ is the number of years (in this case 1 year).
We are given that $P = 3,500$, $r = 0.03$, $n = 1$, and $t = 1$. Substituting these values into the formula, we have:
\[A = 3500\left(1+\frac{0.03}{1}\right)^{1\cdot 1}.\]
Simplifying the expression inside the parentheses gives:
\[A = 3500(1+0.03)^1.\]
Adding the numbers inside the parentheses:
\[A = 3500(1.03)^1.\]
Evaluating the exponent:
\[A = 3500(1.03).\]
Finally, multiplying:
\[A = \boxed{3,605}.\]
Therefore, the amount in the account at the end of one year is $\boxed{\$3,605}$.
Suppose that $3,500 is placed in an account that pays 3% interest compounded each year assume that no withdrawals are made from the account find the amount in the account at the end of one year
1 answer
1 answer