anonymous
  • anonymous
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
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
To find compounding interest use the formula P*e^(rt), where P is the principal, r is the rate, and t is time.
anonymous
  • anonymous
You were probably given a formula for anually compounded interest as being something like \(A(t) = A_0*e^{rt}\)
anonymous
  • anonymous
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).

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anonymous
  • anonymous
And t is the number of years.
amistre64
  • amistre64
ah yes...Pert :)
anonymous
  • anonymous
so would this A(t)=5000xe^.069(30) be correct.. ?
anonymous
  • anonymous
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.
anonymous
  • anonymous
Got it :]
anonymous
  • anonymous
Yes. \(A(t) = 5000 * e^{.069*30} = 5000 * e^{2.07} = 5000 * 7.9248 = 39624.11\)
anonymous
  • anonymous
Thanksss ! :D
anonymous
  • anonymous
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.
anonymous
  • anonymous
Not quite the same. But similiar. You were probably given an equation for exponential decay?
anonymous
  • anonymous
no/: i've just been emailed the assignments . i think you've helped me more than my teacher has -_- .
anonymous
  • anonymous
yuppp pretty much ! =)
anonymous
  • anonymous
Well ok, exponential decay has the form \(A(t) = A_0*2^{\frac{t}{t_{halflife}}}\)
anonymous
  • anonymous
err The exponent on the 2 should be \[\frac{-t}{t_{halflife}}\]
anonymous
  • anonymous
And \(A_0\) is again the initial amount.

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