Question

Elizabeth invested $970 in an account paying an interest rate of 6\tfrac{5}{8}6 8 5 ​ % compounded daily. Matthew invested $970 in an account paying an interest rate of 6\tfrac{3}{4}6 4 3 ​ % compounded continuously. After 8 years, how much more money would Matthew have in his account than Elizabeth, to the nearest dollar?

Answer 1

Using continuous compounding and compound interest, it is found that Matthew would have $17 more than Elizabeth in his account.

Compound interest:

$A(t) = P\left(1 + \frac{r}{n}\right)^{nt}$

Continuous compounding:

$A(t) = Pe^{rt}$

The parameters are:

• A(t) is the amount of money after t years.  
• P is the principal(the initial sum of money).  
• r is the interest rate(as a decimal value).  
• n is the number of times that interest is compounded per year.  
• t is the time in years for which the money is invested or borrowed.

For both of them:

• Investment of $970, hence $P = 970$

• Invested for 8 years, hence $t = 8$

Elizabeth:

• Compounded daily, hence $n = 365$.

• Rate, as a percent, of $6\frac{5}{8} = 6 + \frac{5}{8} = 6.625$, hence $r = 0.06625$.

Then:

$A(t) = P\left(1 + \frac{r}{n}\right)^{nt}$

$A(8) = 970\left(1 + \frac{0.06625}{365}\right)^{365(8)}$

$A(8) = 1648$

Matthew:

• Rate, as a percent, of $6\frac{3}{4} = 6 + \frac{3}{4} = 6.75$, hence $r = 0.0675$.

Then:

$A(t) = Pe^{rt}$

$A(8) = 970e^{0.0675(8)}$

$A(8) = 1665$

The difference is:

1665 - 1648 = 17

Hence, Matthew would have $17 more than Elizabeth in his account.

A similar problem is given at https://brainly.com/question/24507395