Use the compound interest formulas A = P (1+r/n) nt and A = Pert to solve. Suppose that you have $11,000 to invest. Which investment yields the greater return over 10 years: 6.25% compounded continuously or 6.3% compounded semiannually?

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- nthenic_oftime

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- misty1212

HI!!

- nthenic_oftime

hey there how are you :) @misty1212

- misty1212

\(6.25\%\) compounded continuously would give \[\large11,000\times e^{0.0625\times 10}\] or
\[\huge 11,000\times e^{0.625}\]

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## More answers

- misty1212

i am good
you?

- misty1212

for "semiannually" \(n=2\) you get
\[\large 11,000(1+\frac{0.063}{2})^{2\times 10}\] or \[\huge 11,000\left(1+\frac{.063}{2}\right)^{20}\]

- misty1212

got a calculator?

- nthenic_oftime

i sure do right here... could you explain a lil the first step you did sso i can do the same thing with 6.3%?

- misty1212

for the first one "continuously" you use the formula
\[P\times e^{rt}\] where \(t\) is years and \(r\) is the rate of interest WRITTEN AS A DECIMAL

- misty1212

your rate was \(6.25\%\) but you need to write it as a decimal first, so \(6.25\%=0.0625\)

- misty1212

and \(p=11,000\) and also \(t=10\) so we plug those in to the formula

- misty1212

that is where the \[\huge 11,000\times e^{0.625}\] comes from

- nthenic_oftime

okay im following.

- misty1212

now it is a calculator exercise

- nthenic_oftime

so for 6.3% semiannually i got 20453.95 and 20550.70 for the 6.25% okay i got it thank you misty :) i appriciate it.

- nthenic_oftime

medal for you :)

- misty1212

\[\color\magenta\heartsuit\]

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