Jolene invests her savings in two bank accounts, one paying 4 percent and the other paying 11 percent simple interest per year. She puts twice as much in the lower-yielding account because it is less risky. Her annual interest is 4408 dollars. How much did she invest at each rate? 

Jolene invested $46,378.95 at 4 percent and $23,189.47 at 11 percent to achieve her annual interest of $4408 by setting up and solving a system of equations based on her investment strategy.

Jolene invests her savings in two different bank accounts, one with a 4 percent interest rate and another with an 11 percent interest rate. She invests twice as much in the lower-yielding account. To determine how much she invested at each rate given her annual interest of $4408, we can set up a system of equations.

Let's denote the amount invested at 4 percent as x and the amount invested at 11 percent as y. Given that she puts twice as much in the account with 4 percent, we can say that x = 2y. Furthermore, the total interest earned from both accounts is $4408, so the equation for the total interest would be 0.04x + 0.11y = 4408.

Substitute x = 2y into the interest equation to get 0.04(2y) + 0.11y = 4408. After simplification, the equation becomes 0.19y = 4408. Dividing both sides by 0.19 gives us y = 23189.47. Since x = 2y, multiply y by 2 to find x = 46378.95.

Therefore, Jolene invested $46,378.95 at 4 percent and $23,189.47 at 11 percent.

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