Answer:
i)no as 42>33.5
ii)around 82 %
Explanation:
volume of the scoop - 4/3 pi r cube = ~33.5
volume of cone - 1/3 pi r square = ~42
Percent of cone will be filled is 81%
Given that;
Length of sugar cone = 10cm
Diameter of sugar cone = 4 cm
Diameter of spherical scoop = 4 cm
Computation:
Diameter of sugar cone = 4 cm
Radius of sugar cone = 4/2 = 2 cm
Diameter of spherical scoop = 4 cm
Radius of spherical scoop = 4/2 = 2 cm
Volume of cone = [tex]\frac{1}{3}[\pi r^2h][/tex]
Volume of cone = [tex]\frac{1}{3}[(3.14) (2)^2(10)][/tex]
Volume of cone = 41.86 cm³
Volume of spherical scoop = [tex]\frac{4}{3}[\pi r^3][/tex]
Volume of spherical scoop = [tex]\frac{4}{3}[(3.14) (2)^3][/tex]
Volume of spherical scoop = 33.49 cm³
Cone is enough big.
Percent of cone will be filled = [tex]\frac{33.49}{41.86} \times 100[/tex]
Percent of cone will be filled = 81% (Approx.)
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