Tuesday

October 21, 2014

October 21, 2014

Posted by **Jean** on Wednesday, December 5, 2012 at 1:44pm.

- Business Math -
**Henry**, Friday, December 7, 2012 at 7:18pmP1 = 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.

- Business Math -
**Anonymous**, Thursday, December 20, 2012 at 1:29pmαδε

**Answer this Question**

**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 - 1. Lee Holmes deposited $16,700 in a new savings account at 6% ...

Business Math - Shelley Katz deposited $30,000 in a savings account at 5% ...

Math - Lee Holmes deposited $16,600 in a new savings account at 9% interest ...

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 - John Lee's savings account has a balance of $4494. After 3 years, what ...

math,help - what formula do i have to use for this problems. Problem #4 John Lee...