Correct Answer: $3\lt X\lt 9$

Explanation:

$x^{2}-13x+40\ge 0$

=> $x^{2}-8x-5x+40\ge 0$

=> $x(x-8)-5(x-8)\ge 0$

=> (x-8)(x-5)$\ge 0$

Here the product of (x-8) and (x-5) is grater than zero or equal to zero. so the value of these

When \both positive.(x-8)$\ge$0 =>x$\ge$8:(x-5)$\ge$0 => x$\ge$5

When \both positive, -(x-8)$\ge$0 => x$\le$8:

-(x-5)$\ge$0=> x$\le$5:

$3\lt X\lt 9$