Avery invested $2,100 in an account paying an interest rate of 7 7/8% compounded continuously. Morgan invested $2,100 in an account paying a

Question

Avery invested $2,100 in an account paying an interest rate of 7 7/8% compounded continuously. Morgan invested $2,100 in an account paying an interest rate of 8 1/4 % compounded annually. After 12 years, how much more money would Morgan have in her account than Avery, to the nearest dollar?

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Eden 1 week 2021-11-26T04:03:19+00:00 1 Answer 0

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    2021-11-26T04:04:52+00:00

    Answer:

    \$34

    Step-by-step explanation:

    step 1

    Avery

    we know that

    The formula to calculate continuously compounded interest is equal to

    A=P(e)^{rt}  

    where  

    A is the Final Investment Value  

    P is the Principal amount of money to be invested  

    r is the rate of interest in decimal  

    t is Number of Time Periods  

    e is the mathematical constant number

    we have  

    t=12\ years\\ P=\$2,100\\ r=7 7/8\%=7.875\%=0.07875  

    substitute in the formula above  

    A=\$2,100(e)^{0.07875*12}=\$5,402.91

    step 2

    Morgan

    we know that    

    The compound interest formula is equal to  

    A=P(1+\frac{r}{n})^{nt}  

    where  

    A is the Final Investment Value  

    P is the Principal amount of money to be invested  

    r is the rate of interest  in decimal

    t is Number of Time Periods  

    n is the number of times interest is compounded per year

    in this problem we have  

    t=12\ years\\ P=\$2,100\\ r=8 1/4\%=8.25\%=0.0825\\n=1  

    substitute in the formula above  

    A=\$2,100(1+\frac{0.0825}{1})^{1*12}=\$5,436.94  

    step 3

    Find the difference

    \$5,436.94-\$5,402.91=\$34.03

    To the nearest dollar

    \$34.03=\$34

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