## 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

1. anonymous

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

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

3. 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).

4. anonymous

And t is the number of years.

5. amistre64

ah yes...Pert :)

6. anonymous

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

7. 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.

8. anonymous

Got it :]

9. anonymous

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

10. anonymous

Thanksss ! :D

11. 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.

12. anonymous

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

13. anonymous

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

14. anonymous

yuppp pretty much ! =)

15. anonymous

Well ok, exponential decay has the form $$A(t) = A_0*2^{\frac{t}{t_{halflife}}}$$

16. anonymous

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

17. anonymous

And $$A_0$$ is again the initial amount.