Posted by Jean on Wednesday, December 5, 2012 at 1:44pm.
P1 = Po(1+r)^n.
P1 = Principal after 1st 3 years.
Po1 = $15,000 = Initial deposit @ beginning of the 1st 3 years.
r = (9%/2)/100% = 0.045 = Semi-annual %
rate expressed as a decimal.
n = 2Comp/yr * 3yrs = 6 Compounding
periods.
Solve the given Eq and get:
P1 = $19,533.90.
P2 = Po2(1+r)^n.
P2 = Principal amount after the 2nd 3 years.
Po2 = 19,533.90 + $40,000 = $59,533.90
= Initial deposit for 2nd 3 years.
Solve for P2.
Answer: P2 = $77,528.63.
NOTE: r,and n are the same for both P1 and P2 calculations.
αδε
Related Questions
Math - Lee Holmes deposited $15,000 in a new savings account at 9% interest ...
MATH HOMEWORK HELP PLEASE ANSWER - Lee Holmes deposited $15,000 in a new savings...
Math - Lee Holmes deposited $16,600 in a new savings account at 9% interest ...
Business Math - Shelley Katz deposited $30,000 in a savings account at 5% ...
math - Jacob Fonda deposited $25,000 in a savings account at 10% interest ...
math - Jacob Fonda deposited $25,000 in a savings account at 10% interest ...
Math - Lee Holmes deposited $16,600 in a new savings account at 9% interest ...
Business Algebra - Invest $23,000 in a savings account at 4.25% interest ...
math - Suppose that $200 was deposited on 1st Jan 2000 into an account that ...
math - What is the size of the payments that must be deposited at the beginning ...
For Further Reading