You invest $500 in an account with an annual interest rate of 1.1%, compounded continuously. How much money is in the account after 10 years?
A=P(1+\(\frac{r}{100} )^{n}\)
A=500(1+\(\frac{1.1}{100} )^{10}\)
A=500(1.011\()^{10}\)
A=500(1.1156)
A=557.80
For continuous compounding, the formula is:
\(A=Pe^{rt}\)
where e is the base of natural logs.
\(A=500\timese^{0.011\times10}=558.14\)
So the amount after 10 years is $558.14