Respuesta :
Hello
To solve this question, we would need to use the forumla of compound interest
[tex]\begin{gathered} p=10,000 \\ a=11,000 \\ r=12\text{ \%=0.12} \\ t=\text{ ?} \\ n=12 \end{gathered}[/tex]Let's input the data into the forumla
[tex]\begin{gathered} a=p(1+\frac{r}{n})^{nt} \\ 11000=10000(1+\frac{0.12}{12})^{12\times t} \\ \frac{11000}{10000}=(1+0.01)^{12t} \\ 1.1=1.01^{12t} \\ \text{take log of both sides} \\ \log 1.1=12t\log 1.01 \\ 12t=\frac{\log 1.1}{\log 1.01} \\ 12t=9.579 \\ \frac{12t}{12}=\frac{9.579}{12} \\ t=0.798\text{years} \end{gathered}[/tex]This can be converted to months
[tex]\begin{gathered} 12\text{months}=1\text{year} \\ x=0.798 \\ x=12\times0.798 \\ x=9.579\text{ months} \\ x\approx10\text{months} \end{gathered}[/tex]It would take approximately 10 months to get 11,000 from a principal of 10,000 at a rate of 12%
