Answer:
The rate is greater than 8%
Explanation:
Given
[tex]\small I = 1,000 \left (\left (1+\frac{r}{100} \right )^{n}-1 \right )[/tex]
Missing part of question
[tex]I =210[/tex]
[tex]n =2[/tex]
Required
Is r > 1
We have:
[tex]\small I = 1,000 \left (\left (1+\frac{r}{100} \right )^{n}-1 \right )[/tex]
Substitute values for r and I
[tex]210 = 1,000 \left (\left (1+\frac{r}{100} \right )^{2}-1 \right )[/tex]
Divide both sides by 1000
[tex]0.210 = \left (\left (1+\frac{r}{100} \right )^{2}-1 \right )[/tex]
Add 1 to both sides
[tex]1.210 = (1+\frac{r}{100} \right ))^{2}[/tex]
Take square roots of both sides
[tex]\sqrt{1.210} = 1+\frac{r}{100}[/tex]
[tex]1.1 = 1+\frac{r}{100}[/tex]
Subtract 1 from both sides
[tex]0.1 = \frac{r}{100}[/tex]
Multiply both sides by 100
[tex]r = 10[/tex]
[tex]10 > 8[/tex]
Hence, the rate is greater than 8%