a.
Given the equation of the jumps as;
[tex]f(x)=-0.2x^2+1.6x[/tex]We want to find the range of the jump.To do this let us substitute f(x) = 0 and solve for x;
[tex]\begin{gathered} f(x)=-0.2x^2+1.6x=0 \\ -0.2x^2+1.6x=0 \\ -0.2x^2=-1.6x \\ \text{divide both sides by -0.2x} \\ \frac{-0.2x}{-0.2x}^2=\frac{-1.6x}{-0.2x} \\ x=8\text{ ft} \end{gathered}[/tex]Therefore, the mouse can jump 8 ft horizontally.
b.
We want to know if the mouse can jump over a 3ft fence;
Let's substitute x=8/2 =4 into the equation to get the maximum height the mouse can reach;
[tex]\begin{gathered} f(x)=-0.2x^2+1.6x \\ f(4)=-0.2(4)^2+1.6(4) \\ f(4)=3.2ft \end{gathered}[/tex]since the maximum height is more than 3ft, it means the mouse can jump higher than 3 ft.
since the maximum height the mouse can jump is 3.2ft, and the fence is 3 ft, the difference between the maximum height and the height of the fence is;
[tex]\begin{gathered} 3.2ft-3.0ft \\ =0.2ft \end{gathered}[/tex]So, the mouse can jump 0.2ft higher than the 3ft fence.
Therefore, the mouse can clear it with 0.2 ft.
[tex]\text{Yes, the mouse can clear it by 0.2ft}[/tex]