Addison invested $950 in an account paying an interest rate of 1, one eigth1 81 % compounded annually. Alexandra invested $950 in an account paying an interest rate of 0, one half0 21 % compounded daily. After 5 years, how much more money would Addison have in her account than Alexandra, to the nearest dollar?
Question
Addison invested 950 in an account paying an interest rate of 0, one half0 21 % compounded daily. After 5 years, how much more money would Addison have in her account than Alexandra, to the nearest dollar?
Solution
The question seems to have some typographical errors in the interest rates. However, I'll assume that Addison's interest rate is 1.81% compounded annually and Alexandra's interest rate is 0.21% compounded daily.
Here's how you can calculate it:
Step 1: Convert the interest rates from percentages to decimals. So, 1.81% becomes 0.0181 and 0.21% becomes 0.0021.
Step 2: Calculate the amount in Addison's account after 5 years using the formula for compound interest: A = P(1 + r/n)^(nt), where:
- A is the amount of money accumulated after n years, including interest.
- P is the principal amount (the initial amount of money).
- r is the annual interest rate (in decimal).
- n is the number of times that interest is compounded per year.
- t is the time the money is invested for, in years.
For Addison, since the interest is compounded annually, n = 1. So, her account balance would be A = 950(1 + 0.0181/1)^(1*5) = $1,089.47 (rounded to the nearest cent).
Step 3: Calculate the amount in Alexandra's account. Since her interest is compounded daily, n = 365. So, her account balance would be A = 950(1 + 0.0021/365)^(365*5) = $1,010.53 (rounded to the nearest cent).
Step 4: Subtract Alexandra's balance from Addison's to find out how much more money Addison would have. So, 1,010.53 = $78.94.
So, to the nearest dollar, Addison would have $79 more in her account than Alexandra after 5 years.
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