For p = $3000, r = 3.5%,and t = 5 years, find the balance in an account when interest is compounded (a) quarterly, (b) monthly, and (c) continuously.

Step-by-step explanation:

a) P = 3000

It was compounded quarterly. This means that it was compounded 4 times in a year. So

n = 4

The rate at which the principal was compounded is 3.5%. So

r = 3.5/100 = 0.035

It was compounded for just 5 years. So

t = 5

The formula for compound interest is

A = P(1+r/n)^nt

A = total amount in the account at the end of t years. Therefore

A = 3000 (1+0.035/4)^4×5

A = $3571

b) It was compounded monthly. This means that it was compounded 12 times in a year. So

n = 12

Therefore,

A = 3000 (1+0.035/12)^12×5

A = $3572.88

c) The formula for continuously compounded interest is

A = Pe (r x t)

where

e = 2.7183.

Therefore

A = 3000 × 2.7183^(0.035 × 5)

A = $3573.7

set each factor equal to 0

4x-3 = 0

2x-1 = 0

solve for x.

4x-3=0

4x=3

x= 3/4

solve for x. 

2x-1= 0

2x= 1

x= 1/2

your answer should be 1/2 is less than or equal to x and 3/4 is greater than or equal to x