On January 1, 1994, Jill invested P dollars in an account that pays in

Between 1st Jan 1994 & 1st Jan 1998, time elapsed = 4 years

Use the Compound interest formula to calculate the amount

\(A=P(1+\frac{r}{100})^n\) \(=> P(1+\frac{8}{100})^4\)

or \(A=P(1.08)^4\)

Option D

Knowing the formula for compound interest helps a lot here.

Compound interest is given by

\(A = P(1 +\frac{r}{n})^{nt}\)

A = final amount
P = principal invested
r = interest rate in decimal form
n = number of compounding periods per year
t = time

Here interest compounds annually; it earns "interest on interest," and pays one time per year, on December 31.

So r = .08
Because \(n = 1\), \((\frac{.08}{1}) = .08\)

Then \(1 + .08 = 1.08\)

Time t, = 4 years: she gets paid December 31 of 1994, 1995, 1996, and 1997

Where n = 1 and t = 4, thus: \((1.08)^{1*4} = (1.08)^4\)

Finally, multiply by the principal: \(P(1.08)^4\)

Answer D
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Since the investment of P dollars was compounded annually for 4 years, the new value of the original investment is P(1.08)^4.

Answer: D
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The easiest/fastest solution involves plugging the given values into the compound interest formula (as others have done).

However, if you forgot that formula, you can always create a table to arrive at the correct answer.

# of year elapsed | value of investment
0 | P
1 | P(1.08)
2 | P(1.08)(1.08) = P(1.08)²
3 | P(1.08)(1.08)(1.08) = P(1.08)³
4 | P(1.08)(1.08)(1.08)(1.08) = P(1.08)⁴

Answer: D

Cheers,
Brent

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