june has a savings account with an annual simple interest rate of 2.6%. she hopes to gain $6,500 in interest…

june has a savings account with an annual simple interest rate of 2.6%. she hopes to gain $6,500 in interest over a period of eleven years. to accomplish this, june invested $17,801 in the account, but later realized that this was not enough money. to the nearest dollar, how much more money should june have initially invested to reach her goal?\na. $6,335\nb. $3,715\nc. $1,409\nd. $4,926\nplease select the best answer from the choices provided\na\nb\nc\nd

june has a savings account with an annual simple interest rate of 2.6%. she hopes to gain $6,500 in interest over a period of eleven years. to accomplish this, june invested $17,801 in the account, but later realized that this was not enough money. to the nearest dollar, how much more money should june have initially invested to reach her goal?\na. $6,335\nb. $3,715\nc. $1,409\nd. $4,926\nplease select the best answer from the choices provided\na\nb\nc\nd

Answer

Explanation:

Step1: Recall simple interest formula

The formula for simple interest is ( I = P \times r \times t ), where ( I ) is the interest, ( P ) is the principal amount, ( r ) is the annual interest rate (in decimal), and ( t ) is the time in years. We need to find the principal ( P ) required to get ( I = 6500 ) with ( r = 0.026 ) and ( t = 11 ).

Step2: Solve for P

Rearrange the formula to solve for ( P ): ( P=\frac{I}{r\times t} ). Substitute the values: ( P = \frac{6500}{0.026\times11} ). First calculate the denominator: ( 0.026\times11 = 0.286 ). Then ( P=\frac{6500}{0.286}\approx22727.27 ).

Step3: Find the difference

June initially invested ( 17801 ). The additional amount needed is ( 22727.27 - 17801 = 4926.27 ), which rounds to ( 4926 ).

Answer:

D. $4,926