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jcdavi34
 one year ago
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?
jcdavi34
 one year ago
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?

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Hunus
 one year ago
Best ResponseYou've already chosen the best response.0The 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 \[A=P(1+\frac{r}{365})^{365t}\]

Hunus
 one year ago
Best ResponseYou've already chosen the best response.0I got something different. Will you show me your steps?

jcdavi34
 one year ago
Best ResponseYou've already chosen the best response.02265.96(1+r/365)365*12

Hunus
 one year ago
Best ResponseYou've already chosen the best response.0Okay is it 12 or 13 years?

Hunus
 one year ago
Best ResponseYou've already chosen the best response.0no? In your question you put 13 in your step you put 12

Hunus
 one year ago
Best ResponseYou've already chosen the best response.0I know. But I need to know if t = 12 or t = 13

Hunus
 one year ago
Best ResponseYou've already chosen the best response.0\[2265.96=1227.43(1+\frac{r}{365})^{365*13}\] \[\frac{2265.96}{1227.43}=(1+\frac{r}{365})^{365*13}\] \[\log(1.8461)=\log((1+\frac{r}{365})^{365*13})\] \[\log(1.8461)=365*13*\log((1+\frac{r}{365}))\] \[\frac{\log(1.8461)}{365*13}=1.000129=\log(1+\frac{r}{365})\] \[10^{1.000129}=1+\frac{r}{365}\] \[10^{1.000129}1=0.000129=\frac{r}{365}\] \[r=0000129*365=0.0471\]
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