Algebra
posted by Sylvie .
Show how you substitute the values into the formula, then use your calculator.
*Use A = P(1+r/n)nt to find the amount of money in an account after t years, compounded n times per year.
*Use I = Prt to find the amount of simple interest earned after t years
1) If a person invests $5780 in an account that pays 9% interest compounded annually, find the balance after 13 years.
I got A = 5780 (1+0.09/1)13, but my calculator answer doesn't seem right.
2) Find the value of $3500 deposited for 9 years in an account paying 5% annual interest compounded semiannually.
3) Find the value of $1200 deposited for 18 years in an account paying 7% annual interest compounded monthly. I got A = 1200(1+.07/12)216, but again, I don't know how to use the calculator for this.
4) Find the amount of simple interest earned if you invest $9000 at .04% interest for 20 years. Then find the account balance.

1. Your expression is correct, I got 17720.35
(this might seem high, but true. Back in the "olden days" before calculators we used something called the rule of 72. It meant if you have a time and rate which multiplied to something close to 72, your money would double.
e.g. at 8% money would double in appr 9 years (9x8=72)
e.g. 100(1.08)^9 = 199.90
in your case it would have doubled in 8 years to appr. 11560 , which would reach 23120 after 18 years.
So 17720 after 13 years is reasonable.
2. amount = 3500(1 + .05/2)^(9(2))
= 3500(1.025)^18 = 5458.81
3. your expression is correct again, perhaps you don't know how to use your calculator
the magic key is the y^{x} key
e.g. to do 4^3
enter 4
press the y^{x} key
enter 3
press =
you should get 64
so for A = 1200(1+.07/12)^216
here are my keystrokes
.07 รท12
=
+1
=
y^{x}
216
= > at this point you should have 3.5125...
x
3500
=
you should get 12293.89