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
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 $\boxed{\$3,605}$.
To find the amount in the account at the end of one year, we can use the formula for compound interest:
A = P(1 + r/n)^(nt)
Where:
A = the amount in the account at the end of the time period
P = the principal amount (initial deposit)
r = annual interest rate (expressed as a decimal)
n = number of times interest is compounded per year
t = number of years
In this case:
P = $3,500
r = 3% = 0.03 (converted to decimal)
n = 1 (interest compounded annually)
t = 1 year
Substituting these values into the formula:
A = 3500(1 + 0.03/1)^(1*1)
Simplifying:
A = 3500(1 + 0.03)^1
A = 3500(1.03)
A = $3,605
Therefore, the amount in the account at the end of one year is $3,605.