Question

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

Answer 1

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

Answer 2

You were probably given a formula for anually compounded interest as being something like \(A(t) = A_0*e^{rt}\)

Answer 3

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).

Answer 4

And t is the number of years.

Answer 5

ah yes...Pert :)

Answer 6

so would this A(t)=5000xe^.069(30) be correct.. ?

Answer 7

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.

Answer 8

Got it :]

Answer 9

Yes. \(A(t) = 5000 * e^{.069*30} = 5000 * e^{2.07} = 5000 * 7.9248 = 39624.11\)

Answer 10

Thanksss ! :D

Answer 11

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.

Answer 12

Not quite the same. But similiar. You were probably given an equation for exponential decay?

Answer 13

no/: i've just been emailed the assignments .
i think you've helped me more than my teacher has -_- .

Answer 14

yuppp pretty much ! =)

Answer 15

Well ok, exponential decay has the form

\(A(t) = A_0*2^{\frac{t}{t_{halflife}}}\)

Answer 16

err The exponent on the 2 should be \[\frac{-t}{t_{halflife}}\]

Answer 17

And \(A_0\) is again the initial amount.