Answer
$\text{approximately } 3.8 \text{ times}$
Work Step by Step
Substituting $x=3$ in the given function, $
F(x)=0.6\sqrt{49-x^2}
,$ results to
\begin{array}{l}\require{cancel}
F(x)=0.6\sqrt{49-3^2}
\\
F(x)=0.6\sqrt{49-9}
\\
F(x)=0.6\sqrt{40}
\\
F(x)=0.6\sqrt{4\cdot10}
\\
F(x)=0.6\sqrt{(2)^2\cdot10}
\\
F(x)=0.6(2)\sqrt{10}
\\
F(x)=1.2\sqrt{10}
\\
F(x)\approx3.8
.\end{array}
Hence, the demand per week, $F(x),$ of an older released DVD is $
\text{approximately } 3.8 \text{ times}
.$
