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nthenic_oftime

  • one year ago

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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  1. misty1212
    • one year ago
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    HI!!

  2. nthenic_oftime
    • one year ago
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    hey there how are you :) @misty1212

  3. misty1212
    • one year ago
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    \(6.25\%\) compounded continuously would give \[\large11,000\times e^{0.0625\times 10}\] or \[\huge 11,000\times e^{0.625}\]

  4. misty1212
    • one year ago
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    i am good you?

  5. misty1212
    • one year ago
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    for "semiannually" \(n=2\) you get \[\large 11,000(1+\frac{0.063}{2})^{2\times 10}\] or \[\huge 11,000\left(1+\frac{.063}{2}\right)^{20}\]

  6. misty1212
    • one year ago
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    got a calculator?

  7. nthenic_oftime
    • one year ago
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    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%?

  8. misty1212
    • one year ago
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    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

  9. misty1212
    • one year ago
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    your rate was \(6.25\%\) but you need to write it as a decimal first, so \(6.25\%=0.0625\)

  10. misty1212
    • one year ago
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    and \(p=11,000\) and also \(t=10\) so we plug those in to the formula

  11. misty1212
    • one year ago
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    that is where the \[\huge 11,000\times e^{0.625}\] comes from

  12. nthenic_oftime
    • one year ago
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    okay im following.

  13. misty1212
    • one year ago
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    now it is a calculator exercise

  14. nthenic_oftime
    • one year ago
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    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.

  15. nthenic_oftime
    • one year ago
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    medal for you :)

  16. misty1212
    • one year ago
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    \[\color\magenta\heartsuit\]

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