A cone is inverted and filled with water to 3/4 of its height. What percent of the cone's volume is filled with water? Express your answer a

Let's call the height of the cone  " h " and the area of its base  " a ".

volume of cone  =  (1/3)(a)(h)

The water will also be in the shape of a cone, with a height that is 3/4 that of the container.

volume of water  =  (1/3)(a)(3/4 * h)

volume of water  =  (3/4) * (1/3)(a)(h)

volume of water  =  (3/4) * volume of cone

volume of water  =  75% of the volume of cone

75% of the cone's volume is filled with water.

Hectictar's answer is correct if the tank is a cylinder....however....a cone is a little different because of the variable cross-sections

We  can use similar triangles here......call the radius of the tank  r and the height h

So.....if the water is filled to 3/4 of the inverted cone's height.....we have

r / h  =  x / [3/4] h     where x =  the radius of the cone formed at the 3/4 water level

So.....x =  3/4 r

So.....the volume of the water  when the tank is full is   pi * r^2 * h / 3

And the volume of water at the 3/4  level  is   [pi *  [ (3/4) r ] ^2 * [ (3/4)] h ] / 3 =

[ 27/64] pi* r^2 * h / 3

So....the percent of the cone filled with water  is   27/64  ≈ .4219  ≈ 42.2%

Always happy to help out a Bama fan....LOL!!!!!