at an 114417 X b 4 years c 12 years 1713 79 X the nearest ce

at an 1,14417 X (b) 4 years (c) 12 years 1.713 79 X the nearest cent.)

Solution

A = P(1 + r_/n)^nt ,A is the amount of investment after t years.P is the principal amount invested , r is the annual rate of interest, n= number of times compounded in year

8)

P=1000,r=4.5%=0.045,n=365

a)

t=3

A = 1000(1 + (0.045_/365))^365*3

A = 1144.53 dollars

b)

t=4

A = 1000(1 + (0.045_/365))^365*4

A = 1197.20 dollars

c)

t=12

A = 1000(1 + (0.045_/365))^365*12

A =  1715.95 dollars

================================================================

9)

P=1500,r=6.75%=0.0675,n=4

a)

t=1

A = 1500(1 + (0.0675_/4))^4*1

A = 1603.84 dollars

b)

t=2

A = 1500(1 + (0.0675_/4))^4*2

A = 1714.87 dollars

c)

t=8

A = 1500(1 + (0.0675_/4))^4*8

A = 2562.44 dollars