SOLUTION: Aaron invested $7,000 in an account paying an interest rate of 3 5/8 compounded continuously. Mia invested $7,000 in an account paying an interest rate of 3 3/4% compounded monthly

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Question 1159176: Aaron invested $7,000 in an account paying an interest rate of 3 5/8 compounded continuously. Mia invested $7,000 in an account paying an interest rate of 3 3/4% compounded monthly. After 17 years , how much more money would Mia have in her account than aaron, to the nearest dollar?
Answer by ikleyn(52798) About Me  (Show Source):
You can put this solution on YOUR website!

35%2F8% = 0.03625.

Aaron's account after 17 years  A = 7000%2Ae%5E%280.03625%2A17%29 = 7000%2A2.71828%5E%280.03625%2A17%29 = 12963.79 dollars.



33%2F4% = 0.0375.

Mia's account after 17 years M = 7000%2A%281%2B0.0375%2F12%29%5E%2817%2A12%29 = 13229.06 dollars.



Mia - Aaron (the difference) is  13229.06 - 12963.79 = 265.27, or, rounded to the nearest dollar, 265 dollars.    ANSWER

Solved.

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On discretely compounded and continuously compounded accounts see the lessons
    - Problems on discretely compound accounts
    - Problems on continuously compound accounts
in this site.

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    - ALGEBRA-I - YOUR ONLINE TEXTBOOK.

The referred lessons are the part of this online textbook under the topic "Logarithms".


Save the link to this online textbook together with its description

Free of charge online textbook in ALGEBRA-I
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to your archive and use it when it is needed.