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anonymous

  • 5 years ago

I'm having a hard time comprehending this/: Can someone explain ? Find the amount in a continuously compounded account for the given conditions. Principal: $5000 Annual Interest Rate: 6.9% Time: 30yr

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  1. anonymous
    • 5 years ago
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    To find compounding interest use the formula P*e^(rt), where P is the principal, r is the rate, and t is time.

  2. anonymous
    • 5 years ago
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    You were probably given a formula for anually compounded interest as being something like \(A(t) = A_0*e^{rt}\)

  3. anonymous
    • 5 years ago
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    Where \(A_0\) is the principal (or amount you start with) and r is the rate as a decimal number (e.g. 6.9% = .069).

  4. anonymous
    • 5 years ago
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    And t is the number of years.

  5. amistre64
    • 5 years ago
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    ah yes...Pert :)

  6. anonymous
    • 5 years ago
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    so would this A(t)=5000xe^.069(30) be correct.. ?

  7. anonymous
    • 5 years ago
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    Just make sure you multiply the year and rate THEN raise e to that product. Don't accidentally raise e to the rth power then multiply by t. So it would be e^(.069*30), not e^.069*30.

  8. anonymous
    • 5 years ago
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    Got it :]

  9. anonymous
    • 5 years ago
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    Yes. \(A(t) = 5000 * e^{.069*30} = 5000 * e^{2.07} = 5000 * 7.9248 = 39624.11\)

  10. anonymous
    • 5 years ago
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    Thanksss ! :D

  11. anonymous
    • 5 years ago
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    Is this one the same ? Hg-197 is used in kidney scans. It has a half-life of 64.128 h. Write the exponential decay function for a 12-mg sample. Find the amount remaining after 72 h.

  12. anonymous
    • 5 years ago
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    Not quite the same. But similiar. You were probably given an equation for exponential decay?

  13. anonymous
    • 5 years ago
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    no/: i've just been emailed the assignments . i think you've helped me more than my teacher has -_- .

  14. anonymous
    • 5 years ago
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    yuppp pretty much ! =)

  15. anonymous
    • 5 years ago
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    Well ok, exponential decay has the form \(A(t) = A_0*2^{\frac{t}{t_{halflife}}}\)

  16. anonymous
    • 5 years ago
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    err The exponent on the 2 should be \[\frac{-t}{t_{halflife}}\]

  17. anonymous
    • 5 years ago
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    And \(A_0\) is again the initial amount.

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