What would be the compound amount after 19 years on an investment of $42,000 with an 11% interest rate compounded annually?

P = Po(1+r)^n.

Po = $42,000
r = 11%/100% = 0.11
n = 1comp./yr. * 19yrs = 19 Compounding
periods.

Plug the above values into the given Eq and solve for P.

See Related Questions: Mon,7-1-13,10:

Ok totally lost here. For someone new to this could you please explain in understandable terms?

To calculate the compound amount after 19 years on an investment, you can use the formula for compound interest:

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

Where:
A = compound amount
P = principal amount (initial investment)
r = annual interest rate (as a decimal)
n = number of times interest is compounded per year
t = number of years

In this case, the principal amount (P) is $42,000, the annual interest rate (r) is 11% (or 0.11 as a decimal), and the interest is compounded annually, so n = 1. The investment period (t) is 19 years.

Using the given values, we can calculate the compound amount (A):

A = $42,000(1 + 0.11/1)^(1*19)

Simplifying the equation:

A = $42,000(1 + 0.11)^19

A = $42,000(1.11)^19

Calculating the exponent:

A = $42,000(3.483055848377673)

A ≈ $146,437.32

Therefore, the compound amount after 19 years on an investment of $42,000 with an 11% interest rate compounded annually would be approximately $146,437.32.