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Sagot :
[tex]r\approx0.139[/tex]
1) Since this is a Continuously Compounded operation in a 10 yrs period, then we can write out the following equation:
[tex]\begin{gathered} A=Pe^{rt} \\ \end{gathered}[/tex]2) Plugging into the equation the given data and since Otto is 20 yrs old and he plans to get $4,000 in ten years, we can write out:
[tex]\begin{gathered} 4000=1000e^{\mleft\{10r\mright\}} \\ \frac{4000}{1000}=\frac{1000e^{10r}}{1000} \\ 4=e^{10r} \\ \ln (4)=\ln (e)^{10r} \\ 10r=\ln (4) \\ r=\frac{\ln (4)}{10} \\ r=0.1386 \end{gathered}[/tex]3) Thus the rate Otto needs is
[tex]r=0.1386\approx0.139[/tex]Note that since the 0.1386 the six here is greater than 5 then we can round up to the next thousandth, in this case: 0.139. For the 0.1386 is closer to 0.139 (0.004) than to 0.138 (0.006).
Or 13.9%
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