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Let V be the volume of the ice cube and U be the volume of the cube immersed in water

density of water at 4oC = 0.998 g/cm³

density of ice at 0oC = 0.917 g/cm³

Weight of the ice cube = volume * density * g = 0.917*V*g [N]

Buoyancy on the ice cube = volume * density * g = 0.988*U*g [N]

Apply Newton's 3rd Law of Motion to the floating ice cube:

0.917*V*g = 0.988*U*g

U/V = 0.928 = 92.8%

Hence, 92.8% of the ice cube is immersed in water, or 7.2% of the ice cube is above water. The answer in percent can be converted to a fraction as follows: 7.2/100.

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