Answer:
Given the statement:
8,000 earn in four years compounded daily at 5%
To find the amount we use formula:
[tex]A = P(1+\frac{r}{n})^{nt}[/tex]
where P is the principal , A is the amount , n is number of times compounded per year and t is the time in year.
Here, Principal(P) = $8000, r = 5% and n = 365
Substitute these given values we get;
[tex]A_1= 8000(1+\frac{5}{365})^{365 \cdot 4}[/tex]
[tex]A_1= 8000 \cdot 1.000137^{1460}[/tex]
[tex]A_1= 8000 \cdot 1.22141[/tex]
Simplify:
[tex]A_1= \$9771.28[/tex]
To find the Interest we use formula:
[tex]I_1= A_1-P[/tex]
[tex]I_1= 9771.28 -8000 = \$1771.28[/tex]
It is also given that:
8,000 earn in four years compounded annually at 5%.
Here, P = $8000, r = 5% , t =4 year and n = 1
Using the same formula to calculate the amount:
[tex]A_2 = 8000(1+\frac{5}{1})^{1 \cdot 4}[/tex]
[tex]A_2= 8000(1.05)^4[/tex]
Simplify:
[tex]A_2= \$9724.05[/tex]
To find the Interest :
[tex]I_2= A_2 - P[/tex]
[tex]I_2= 9724.05 - 8000= \$1724.05[/tex]
Then;
[tex]I_1-I_2 = 1771.28-1724.05 = \$47.23[/tex]
Therefore, $47.23 more would $8,000 earn in four years compounded daily at 5% than compounded annually at 5%
