When $70 is invested monthly with an annual compound interest rate of 12% interest, compounded monthly, what is the total amount of the investment at the end of the second month? Round the answer to two decimal places as needed.(1 point) The total amount of the investment at the end of the second month is $
To solve this problem, we can use the formula for compound interest:
\[A = P \left(1 + \frac{r}{n}\right)^{nt}\]
Where:
- $A$ is the total amount of the investment at the end of the specified time period
- $P$ is the principal amount (the initial investment)
- $r$ is the annual interest rate (expressed as a decimal)
- $n$ is the number of times interest is compounded per year
- $t$ is the number of years
In this case, the principal amount is $70, the annual interest rate is 12% or 0.12, the interest is compounded monthly (so $n = 12$), and we are looking at a two-month period (so $t=2/12 = 1/6$ years).
Plugging in these values, we get:
\[A = 70 \left(1 + \frac{0.12}{12}\right)^{(1/6)(12)}\]
Simplifying:
\[A = 70 \left(1 + \frac{0.01}{1}\right)^{2}\]
\[A = 70(1.01)^{2}\]
\[A = 70(1.0201)\]
\[A = 70.41\]
Therefore, the total amount of the investment at the end of the second month is $70.41.
70.41 is wrong
Apologies for the mistake.
To find the total amount of the investment at the end of the second month, we need to consider the interest earned from the first month as well.
In the first month, the investment of $70 earns interest of $70 * (0.12/12) = $0.70.
So at the beginning of the second month, the principal amount is now $70 + $0.70 = $70.70.
Then, using the compound interest formula, the total amount at the end of the second month is:
\[A = 70.70 \left(1 + \frac{0.12}{12}\right)^{(1/12)(2)}\]
\[A = 70.70 \left(1 + \frac{0.01}{1}\right)^{2/12}\]
\[A = 70.70(1.01)^{1/6}\]
Calculating this value approximately, we find:
\[A \approx 70.70(1.004368)\]
\[A \approx 70.99\]
Therefore, the total amount of the investment at the end of the second month is approximately $70.99.