The radius of a sphere is measured to be $$(7.50 \pm 0.85)$$ cm. Suppose the percentage error in its volume is $$x$$. The value of $$x$$, to the nearest integer x, is ___.

The percentage error in the volume of a sphere is related to the percentage error in its radius. Since $$V = \frac{4}{3}\pi r^3$$, the relative error in volume is three times the relative error in radius.

The percentage error in the radius is $$\frac{\Delta r}{r} \times 100 = \frac{0.85}{7.50} \times 100 = 11.333\%$$.

Therefore, the percentage error in volume is $$x = 3 \times 11.333 = 34\%$$.

To the nearest integer, $$x = 34$$.