Question 1113248
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You haven't given any actual measurements, so we can't get a numerical answer.  But it's easy to get an answer as a fraction of the volume of the cylinder.<br>
If we let R be the "big" radius of the cylinder and r be the "small" radius of the cone, then we know r = R/2.<br>
The fact that the cone is "tilted" makes no difference in its volume; it is still
{{{V = (1/3)(pi)(r^2)(h) = (1/3)(pi)((R/2)^2)(h) = (1/12)(pi)(R^2)(h)}}}<br>
So the volume of the cone is 1/12 the volume of the cylinder; that means the space remaining in the cylinder is 11/12 of its total volume.