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fredd [130]
3 years ago
5

Alan deposited $300 in an account that pays 6% interest compounded continuously. Approximately how long will it take for Alan’s

money to triple?
(Use formula A=Pe^rt where A is the accumulation amount, P is the initial amount, r is the annual rate of interest, and t is the elapsed time in years.)

Show your work for credit
Mathematics
1 answer:
Lelechka [254]3 years ago
3 0

9514 1404 393

Answer:

  18.3 years

Step-by-step explanation:

You want ...

  A/P = 3 = e^(rt) . . . for r = 0.06

Taking the natural log, this gives ...

  ln(3) = 0.06t

  t = ln(3)/0.06 ≈ 18.31

It will take about 18.3 years for the value to triple.

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The correct option is

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Read 2 more answers
I don’t understand, HELPPP
goldenfox [79]
\frac{-6xy^{3}z^5}{-36x^{2}y^{2}z^4}

First cross out the -6 and -36. Six will go into 6 one time and 6 will go into -36 six times so we have the following after crossing out.

\frac{-6xy^{3}z^5}{-36x^{2}y^{2}z^4}
\frac{xy^{3}z^5}{6x^{2}y^{2}z^4} <---After crossing out

Next cross out the x / x^2 which will give us the following:
\frac{xy^{3}z^5}{6x^{2}y^{2}z^4} 
\frac{y^{3}z^5}{6xy^{2}z^4} <---After crossing out

Now cross out the y^2 / y^3
\frac{y^{3}z^5}{6xy^{2}z^4}
\frac{yz^5}{6xz^4} <----After crossing out

All that is left is to cross out the z^5 / z^4 and we are done.
\frac{yz^5}{6xz^4}
\frac{yz}{6x} <----After crossing out and we are done.

\frac{-6xy^{3}z^5}{-36x^{2}y^{2}z^4} = \frac{yz}{6x}





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