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Abigail has an account that pays 6.92% simple interest per year and wants to accumulate $5,896 in interest from this account over six years. How much money should Abigail invest in this account to meet this goal? a. $2,448.02 b. $8,344.10 c. $14,200.39 d. $20,096.67

Respuesta :

[tex]\bf \qquad \textit{Simple Interest Earned}\\\\ I = Prt\qquad \begin{cases} I=\textit{interest earned}\to &\$5896\\ P=\textit{original amount deposited}\\ r=rate\to 6.92\%\to \frac{6.92}{100}\to &0.0692\\ t=years\to &6 \end{cases} \\\\\\ 5896=P\cdot 0.0692\cdot 6[/tex]

solve for P
calculista

we know that

The simple interest formula is equal to

[tex]I=Prt[/tex]

[tex]P=I/(rt)[/tex]

where

I is the amount of money in interest

P is the Principal amount of money to be invested

r is the rate of interest  in decimal

t is Number of Time Periods

in this problem we have

[tex]t=6\ years\\ I=\$5,896\\ P=?\\r=0.0692[/tex]

substitute in the formula above

[tex]P=5,896/(0.0692*6)=\$14,200.39[/tex]

therefore

the answer is

[tex]\$14,200.39[/tex]