(a) [tex]A = 8500(1+\frac{2}{36500})^3^6^5[/tex]
(b) A = $12680.37
(c) [tex]2 = (1.020201)^t[/tex]
Explanation:
Given:
Principal, P = $8,500
Rate of interest, r = 2%
Compounded continuously, n = 365
(a)
Time, t = 1
Amount, A equation = ?
We know:
[tex]A = P(1 + \frac{r}{100 X n} )^n^t[/tex]
On substituting the value we get:
[tex]A = 8500(1+ \frac{2}{100 X 365})^3^6^5 ^X ^1\\ \\A = 8500(1+\frac{2}{36500})^3^6^5[/tex]
(b)
Amount after 20 years = ?
t = 20
[tex]A = 8500(1+ \frac{2}{100 X 365})^3^6^5 ^X ^2^0\\ \\A = 8500(1+\frac{2}{36500})^7^3^0^0\\\\\A = 12680.37[/tex]
(c)
Amount to double = 8500 X 2
               = $17000
Time, t = ?
[tex]17000 = 8500(1+ \frac{2}{100 X 365})^3^6^5 ^X ^t\\ \\2 = (1+\frac{2}{36500})^3^6^5^t\\\\2 = (1.020201)^t[/tex]
