Question

Allie currently has an account balance of $2,265.96. She opened the account 13 years ago with a deposit of $1,227.43. If the interest compounds daily, what is the interest rate on the account?

Answer 1

The formula for compounded interest is
\[A=P(1+\frac{r}{n})^{nt}\]
Where n is the number of times per year it is compounded

So the equation you will need to use is
\[A=P(1+\frac{r}{365})^{365t}\]

Answer 2

i got 1.9%

Answer 3

I got something different. Will you show me your steps?

Answer 4

i got 4.7%

Answer 5

2265.96(1+r/365)365*12

Answer 6

...........

Answer 7

Okay

is it 12 or 13 years?

Answer 8

no

Answer 9

intrest rate

Answer 10

no?

In your question you put 13 in your step you put 12

Answer 11

I know. But I need to know if t = 12 or t = 13

Answer 12

whats r

Answer 13

\[2265.96=1227.43(1+\frac{r}{365})^{365*13}\]
\[\frac{2265.96}{1227.43}=(1+\frac{r}{365})^{365*13}\]
\[\log(1.8461)=\log((1+\frac{r}{365})^{365*13})\]
\[\log(1.8461)=365*13*\log((1+\frac{r}{365}))\]
\[\frac{\log(1.8461)}{365*13}=1.000129=\log(1+\frac{r}{365})\]
\[10^{1.000129}=1+\frac{r}{365}\]
\[10^{1.000129}-1=0.000129=\frac{r}{365}\]
\[r=0000129*365=0.0471\]