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jcdavi34

  • 2 years 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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  1. Hunus
    • 2 years ago
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    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 \[A=P(1+\frac{r}{365})^{365t}\]

  2. jcdavi34
    • 2 years ago
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    i got 1.9%

  3. Hunus
    • 2 years ago
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    I got something different. Will you show me your steps?

  4. jcdavi34
    • 2 years ago
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    i got 4.7%

  5. jcdavi34
    • 2 years ago
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    2265.96(1+r/365)365*12

  6. jcdavi34
    • 2 years ago
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    ...........

  7. Hunus
    • 2 years ago
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    Okay is it 12 or 13 years?

  8. jcdavi34
    • 2 years ago
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    no

  9. jcdavi34
    • 2 years ago
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    intrest rate

  10. Hunus
    • 2 years ago
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    no? In your question you put 13 in your step you put 12

  11. Hunus
    • 2 years ago
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    I know. But I need to know if t = 12 or t = 13

  12. jcdavi34
    • 2 years ago
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    whats r

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

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