Suppose $13,000 is deposited into a savings account earning 3% interest compounded semi-annually. Find the balance in the account after 20 years. Round to the nearest cent.
Use one of these formulas to solve the problem.
We can use the formula for compound interest:
\[A = P \left(1 + \frac{r}{n}\right)^{nt}\]
where $A$ is the final amount, $P$ is the principal (initial amount), $r$ is the annual interest rate (expressed as a decimal), $n$ is the number of times that interest is compounded per year, and $t$ is the number of years.
In this problem, we have $P = 13,000$, $r = 0.03$, $n = 2$ (since interest is compounded semi-annually), and $t = 20$.
Plugging these values into the formula, we get:
\[A = 13000 \left(1 + \frac{0.03}{2}\right)^{(2)(20)}\]
Simplifying the exponent:
\[A = 13000 \left(1 + 0.015\right)^{40}\]
Carrying out the addition inside the parentheses:
\[A = 13000 \left(1.015\right)^{40}\]
Evaluating the expression inside the parentheses:
\[A = 13000 (1.7474220513)\]
Rounding to the nearest cent, we find $A \approx \boxed{22,716.47}$.
To solve this problem, we can use the formula for compound interest:
A = P(1 + r/n)^(nt)
Where:
A = the final amount or balance in the account
P = the initial principal or amount deposited ($13,000)
r = the annual interest rate (3% or 0.03)
n = the number of times interest is compounded per year (2 times, semi-annually)
t = the number of years (20 years)
Plugging in the given values, we get:
A = 13000(1 + 0.03/2)^(2*20)
= 13000(1 + 0.015)^(40)
= 13000(1.015)^(40)
Using a calculator, we can evaluate this expression to get:
A ≈ $28,206.32
Therefore, the balance in the account after 20 years would be approximately $28,206.32.