# $3,000 is invested in an account paying 4% interest and twice that amount is invested at 6%. How do you write the expression that represents the amount of interest earned on the money invested at 6%? ##### 1 Answer Aug 1, 2018 See explanation #### Explanation: Assumption: The interest given is the $\underline{\text{yearly interest rate}}$Twice the amount is $6000

Let the count of years be $n$

$\textcolor{b r o w n}{\text{At 6% "ul("simple")" interest calculated once per year is:}}$

$6000+[nxx(6/100xx$6000)]

Factoring out the $6000 giving: $6000(1+(6n)/100)

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$\textcolor{b r o w n}{\text{At 6% annually "ul("compounded monthly}}$

The 6% is 'spread' over each calculation cycle. There are 12 months in a year so instead of $\frac{6}{100}$ we have $\frac{6}{100} \times \frac{1}{12}$

Note that the count in years is $n$. However ther are 12 calculations cycles in one year so the 'power' of $n$ becomes $12 n$ instead.

$6000(1+6/(12xx100))^(12xxn) $6000(1+cancel(6)^1/cancel(1200)^600)^(12n)

\$6000(1+1/600)^(12n)