Gabriel invested $77,000 in an account paying an interest rate of 3 1/2% compounded monthly. Isabella invested $77,000 in an account paying an interest rate of 4 1/8% compounded quarterly. After 6 years, how much more money would Isabella have in her account than Gabriel, to the nearest dollar?

Respuesta :

WE know, compound interest is given by :

[tex]A = P(1 + \frac{r}{n})^{nt}[/tex]

Here, P is principle amount.

r is annual interest rate.

n, number of times that interest is compounded per unit t .

t, time the money is invested or borrowed for.

To Find :

How much more money would Isabella have in her account than Gabriel, to the nearest dollar.

Solution :

[tex]D = P(1 + \frac{r_1}{n})^{nt}- P(1 + \frac{r_2}{n})^{nt}\\\\D = 77000\times ( 1 + \dfrac{33}{8\times 12})^{12\times 6}-77000\times ( 1 + \dfrac{7}{2\times 12})^{12\times 6}\\\\D = \$(1.257\times 10^{14})[/tex]

Hence, this is the required solution.

Answer: 3534

3534 in delta math

Step-by-step explanation: