Question 764362
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ A\ =\ P\left(1\ +\ \frac{r}{n}\right)^{nt}]


Where *[tex \LARGE A] is the future value, *[tex \LARGE P] is the present value, *[tex \LARGE r] is the interest rate expressed as a decimal, *[tex \LARGE n] is the number of compounding periods per year, and *[tex \LARGE t] is the number of years.


For your situation, you don't care about the actual values of *[tex \LARGE A] and *[tex \LARGE P], just that the ratio is 2:1.  Your *[tex \LARGE t\ =\ 23], and your *[tex \LARGE n\ =\ 1] for annual compounding.  So:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(1\ +\ r\right)^{23}\ =\ 2]


Solve for *[tex \LARGE r]


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
<font face="Math1" size="+2">Egw to Beta kai to Sigma</font>
My calculator said it, I believe it, that settles it
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