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jcdavi34
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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?
 one year ago
 one year ago
jcdavi34 Group Title
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?
 one year ago
 one year ago

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

jcdavi34 Group TitleBest ResponseYou've already chosen the best response.0
i got 1.9%
 one year ago

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

jcdavi34 Group TitleBest ResponseYou've already chosen the best response.0
i got 4.7%
 one year ago

jcdavi34 Group TitleBest ResponseYou've already chosen the best response.0
2265.96(1+r/365)365*12
 one year ago

jcdavi34 Group TitleBest ResponseYou've already chosen the best response.0
...........
 one year ago

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

jcdavi34 Group TitleBest ResponseYou've already chosen the best response.0
intrest rate
 one year ago

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

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

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