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At the beginning of year 1, paolo invests $500

At the beginning of year 1, paolo invests $500-example-1
User Tzerb
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2 Answers

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\bf ~~~~~~ \textit{Compound Interest Earned Amount} \\\\ A=P\left(1+(r)/(n)\right)^(nt) \quad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\to &\$500\\ r=rate\to 4\%\to (4)/(100)\to &0.04\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{annuall, thus once} \end{array}\to &1\\ t=years\to &t \end{cases} \\\\\\ A=500\left(1+(0.04)/(1)\right)^(1\cdot t)\implies A=500(1+0.04)^t \\\\\\ \textit{after 5 years }t=5\qquad A=500(1.04)^5

the example on your picture uses A(n) and n = years, but is pretty much the same, in this case is t = years.
User MariusBudin
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4 votes

Answer:

The correct option is C.

Explanation:

Principal Amount = $500

Rate of Interest = 4%

= 0.04

Time = 4 years

n = 1


Amount = 500*(1+0.04)^(4)\\\\\implies Amount = 500* 1.1699\\\\\implies Amount = \$584.93

Hence, The explicit formula which can be used to find the amount of the money in the account :


A(n)=500\cdot (1+0.04)^((n-1))

Therefore, The correct option is C.

User Zeekvfu
by
8.0k points
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