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 2 years
To find the amount in the account at the end of 2 years, we can use the formula for compound interest:
\[A = P \left(1 + \frac{r}{n}\right)^{nt}\]
Where:
$A$ represents the final amount in the account,
$P$ represents the principal (initial amount), which is $3,500,
$r$ represents the annual interest rate (as a decimal), which is 3% or 0.03,
$n$ represents the number of times interest is compounded per year, which is 1 (compounded annually),
and $t$ represents the number of years, which is 2.
Plugging in the given values, we have:
\[A = 3500 \left(1 + \frac{0.03}{1}\right)^{1 \cdot 2}\]
Simplifying,
\[A = 3500 \left(1 + 0.03\right)^2\]
Evaluating the expression inside the parentheses first,
\[A = 3500 \times 1.03^2\]
Calculating $1.03^2$,
\[A = 3500 \times 1.0609\]
Multiplying,
\[A \approx 3703.15\]
Therefore, the amount in the account at the end of 2 years is approximately $\boxed{\$3703.15}$.
To find the amount in the account at the end of 2 years, we can use the formula for compound interest:
A = P(1 + r/n)^(nt)
Where:
A = the final amount (what we are trying to find)
P = the principal amount (initial deposit), which is $3,500 in this case
r = the annual interest rate in decimal form, which is 3% or 0.03
n = the number of times the interest is compounded per year, in this case, it is compounded once per year (n = 1)
t = the number of years, which is 2
Plugging in the values, we have:
A = 3500(1 + 0.03/1)^(1*2)
A = 3500(1 + 0.03)^2
A = 3500(1.03)^2
A = 3500(1.0609)
A ≈ $3,725.32
Therefore, the amount in the account at the end of 2 years will be approximately $3,725.32.