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)?
We can solve this problem using the formula for compound interest:
\[A = P\left(1 + \frac{r}{n}\right)^{n\cdot t}\]
Where:
- $A$ is the total amount of the investment at the end of the second month
- $P$ is the principal amount, which is $70
- $r$ is the annual interest rate, which is 12%
- $n$ is the number of times the interest is compounded per year, which is monthly and therefore 12
- $t$ is the total investment period in years, which is 2 months and therefore $\frac{2}{12}$
Plugging in the values, we have:
\[A = 70\left(1 + \frac{0.12}{12}\right)^{12\cdot \frac{2}{12}}\]
Simplifying within the parentheses:
\[A = 70\left(1 + \frac{0.01}{12}\right)^{2}\]
\[A = 70\left(1 + 0.000833\right)^{2}\]
And finally, calculating the power:
\[A = 70(1.000833)^{2}\]
\[A \approx 70(1.001666)\]
\[A \approx \boxed{70.12}\]
thats incorrect
I apologize for the mistake in my previous response. Let's go through the solution again.
The formula for compound interest is:
\[A = P\left(1 + \frac{r}{n}\right)^{n\cdot t}\]
Where:
- $A$ is the total amount of the investment at the end of the second month
- $P$ is the principal amount, which is $70
- $r$ is the annual interest rate, which is 12% or 0.12
- $n$ is the number of times the interest is compounded per year, which is monthly and therefore 12
- $t$ is the total investment period in years, which is 2 months and therefore $\frac{2}{12}$
Using these values, we have:
\[A = 70\left(1 + \frac{0.12}{12}\right)^{12\cdot \frac{2}{12}}\]
\[A = 70\left(1 + 0.01\right)^{2}\]
\[A = 70(1.01)^{2}\]
\[A = 70(1.0201)\]
\[A \approx \boxed{71.41}\]
Therefore, the total amount of the investment at the end of the second month is approximately $71.41.
wrong
I apologize for the mistake in my previous responses. Let's go through the solution again.
The formula for compound interest is:
\[A = P\left(1 + \frac{r}{n}\right)^{n\cdot t}\]
Where:
- $A$ is the total amount of the investment at the end of the second month
- $P$ is the principal amount, which is $70
- $r$ is the annual interest rate, which is 12% or 0.12
- $n$ is the number of times the interest is compounded per year, which is monthly and therefore 12
- $t$ is the total investment period in years, which is 2 months and therefore $\frac{2}{12}$
Using these values, we have:
\[A = 70\left(1 + \frac{0.12}{12}\right)^{12\cdot \frac{2}{12}}\]
\[A = 70\left(1 + \frac{0.01}{12}\right)^{2}\]
\[A = 70\left(1 + 0.000833\right)^{2}\]
\[A = 70(1.000833)^{2}\]
\[A \approx 70(1.001666)\]
\[A \approx \boxed{70.12}\]
Therefore, the total amount of the investment at the end of the second month is approximately $70.12.
wrong again
I apologize for the errors in my previous responses. Let's try solving the problem again.
The formula for compound interest is:
\[A = P \left(1 + \frac{r}{n}\right)^{n \cdot t}\]
Where:
- $A$ is the total amount of the investment at the end of the second month
- $P$ is the principal amount, which is $70
- $r$ is the annual interest rate, which is 12% or 0.12
- $n$ is the number of times the interest is compounded per year, which is monthly and therefore 12
- $t$ is the total investment period in years, which is 2 months and therefore $\frac{2}{12}$
Using these values, we have:
\[A = 70 \left(1 + \frac{0.12}{12}\right)^{12 \cdot \frac{2}{12}}\]
Simplifying within the parentheses:
\[A = 70 \left(1 + 0.01\right)^{2}\]
\[A = 70 \cdot 1.01^{2}\]
Calculating the power:
\[A = 70 \cdot 1.0201\]
\[A \approx \boxed{71.41}\]
Therefore, the total amount of the investment at the end of the second month is approximately $71.41.