The volume of a cone is:
[tex]V_{c}=\frac{1}{3}\pi r^2 h \\ \\ \\ Where: \\ \\ r:\text{Radius of the cone} \\ \\ h:\text{height of the cone}[/tex]
Here:
[tex]r=1.75in \\ \\ h=3.5in[/tex]
So, substituting into the formula:
[tex]V_{c}=\frac{1}{3}\pi (1.75)^2(3.5)\approx 11.22in[/tex]
The gum ball is spherical. so the volume of a sphere is:
[tex]V_{e}=\frac{4}{3}\pi r^3 \\ \\ \\ Where: \\ \\ r:\text{Radius of the gum ball} \\ \\ \\ \text{We know that the diameter of this ball is} \ d=0.5in \ \text{so the radius is:} \\ \\ r=0.5/2=0.25in[/tex]
Therefore:
[tex]V_{e}=\frac{4}{3}\pi (0.25)^3\approx 0.065[/tex]
Finally, the volume (V) of the cone that can be filled with flavored ice is:
[tex]V=V_{c}-V_{e}=11.22-0.065 \\ \\ \boxed{V\approx 11.15in^3}[/tex]
