Answer:
Explanation:
Let's first find the total volume of clay used for the cone and the sphere:
For the cone, we need the formula for the volume of a cone: Vcone = (1/3)πr^2h, where r is the radius of the base and h is the height.
Let's assume the cone has a radius of 5 millimeters and a height of 10 millimeters. Then, the volume of the cone is:
Vcone = (1/3)π(5^2)(10) = (1/3)π(250) = 83.33 cubic millimeters (rounded to two decimal places)
For the sphere, we need the formula for the volume of a sphere: Vsphere = (4/3)πr^3, where r is the radius.
Let's assume the sphere has a radius of 4 millimeters. Then, the volume of the sphere is:
Vsphere = (4/3)π(4^3) = (4/3)π(64) = 268.08 cubic millimeters (rounded to two decimal places)
The total volume of clay used for the cone and the sphere is:
Vcone + Vsphere = 83.33 + 268.08 = 351.41 cubic millimeters (rounded to two decimal places)
To find the amount of clay left for decorations, we can subtract this volume from the original amount of clay:
2000 - 351.41 = 1648.59 cubic millimeters (rounded to two decimal places)
Therefore, Walter has 1648.59 cubic millimeters of clay left for decorations.