Use the formula for the volume of a sphere to find the volume of a basketball that has a diameter of 9.2 inches. Use 3.14 as an approximation for pi and round your answer to the nearest hundredth, if necessary. (1 point)
• 407.51 in.'
• 115.55 in.3
• 3, 260.11 in.*
O 1, 222.54 in.3
The formula for the volume of a sphere is given by V = (4/3) * pi * r^3, where V is the volume, pi is approximately 3.14, and r is the radius of the sphere.
Since the diameter of the basketball is 9.2 inches, the radius is half of that, which is 4.6 inches.
Using the formula, we can find the volume of the basketball:
V = (4/3) * 3.14 * (4.6^3)
= (4/3) * 3.14 * (97.336)
= 407.51 in.^3 (rounded to the nearest hundredth)
Therefore, the volume of the basketball is approximately 407.51 in.^3, so the correct answer is:
• 407.51 in.^3
Use the formula for the volume of a sphere to find the volume of a basketball that has a diameter of 9.2 inches. Use 3.14 as an approximation for pi and round your answer to the nearest hundredth, if necessary. (1 point)
• 407.51 in.'
• 115.55 in.3
• 3, 260.11 in.*
O 1, 222.54 in.3
The formula for the volume of a sphere is given by V = (4/3) * pi * r^3, where V is the volume, pi is approximately 3.14, and r is the radius of the sphere.
Since the diameter of the basketball is 9.2 inches, the radius is half of that, which is 4.6 inches.
Using the formula, we can find the volume of the basketball:
V = (4/3) * 3.14 * (4.6^3)
= (4/3) * 3.14 * (97.336)
= 407.516 (rounded to the nearest hundredth)
Therefore, the volume of the basketball is approximately 407.51 in.³, so the correct answer is:
• 407.51 in.³
Find the volume of a sphere with a radius of 1.8 yards. Leave your answer in terms of pi. Round your answer to the nearest hundredth, if necessary. (1 point)
• 23.347 yds®
• 7.2т yds
• 7.787 yds3
• 24.43 yds
The formula for the volume of a sphere is given by V = (4/3) * pi * r^3, where V is the volume, pi is a constant, and r is the radius of the sphere.
Given that the radius of the sphere is 1.8 yards, we can plug this value into the formula:
V = (4/3) * pi * (1.8^3)
= (4/3) * pi * 5.832
≈ 7.787 yds³ (rounded to the nearest hundredth)
Therefore, the volume of the sphere is approximately 7.787 yds³, so the correct answer is:
• 7.787 yds³
Use the formula for the volume of a sphere to solve the following problem.
How much air is needed to fill a basketball with a diameter of 9 inches? Use 3.14 as an approximation for pi. Round your answer to the nearest tenth.
(1 point)
• 190.8 in. 3
• 113 in. 3
• 3, 052.1 in. 8
• 381.5 in. 3
The formula for the volume of a sphere is given by V = (4/3) * pi * r^3, where V is the volume, pi is approximately 3.14, and r is the radius of the sphere.
Since the diameter of the basketball is 9 inches, the radius is half of that, which is 4.5 inches.
Using the formula, we can find the volume of the basketball:
V = (4/3) * 3.14 * (4.5^3)
= (4/3) * 3.14 * (91.125)
≈ 381.511 in.^3 (rounded to the nearest tenth)
Therefore, approximately 381.5 in.^3 of air is needed to fill the basketball, so the correct answer is:
• 381.5 in.^3
The volume of a container measures 4, 000 mm" and contains 50 marbles. Each marble has a radius of 2 mm. Use 3.14 as an approximation for pi. How much space in the box is unused? Round your answer to the nearest tenth.
(1 point)
• 1,674.7 mm*
• 3, 900 mm*
• 33.5 mm
° 2, 325.3 mm
The volume of the box is given as 4,000 mm³, and the radius of each marble is 2 mm.
We can calculate the total volume occupied by the 50 marbles and subtract it from the volume of the box to find the unused space.
The volume of a sphere is given by V = (4/3) * pi * r^3, where V is the volume, pi is approximately 3.14, and r is the radius.
The volume of one marble is V_marble = (4/3) * 3.14 * 2^3
= (4/3) * 3.14 * 8
= 33.493 mm³ (rounded to the nearest thousandth)
The total volume occupied by the 50 marbles is 50 * V_marble = 50 * 33.493 ≈ 1,674.65 mm³
The unused space in the box is then the volume of the box minus the volume of the marbles:
Unused space = Volume of box - Total volume of marbles
= 4,000 - 1,674.65
≈ 2,325.35 mm³ (rounded to the nearest tenth)
Therefore, the correct answer is:
• 2,325.3 mm³
Kairay created a sculpture by forming a hemisphere with a radius of 20 inches and then removing a hemisphere with a radius of 15 inches from it. Calculate the volume of his sculpture. Use 3.14 as an approximation for pi. Round your answer to the nearest whole number. (1 point)
• 7,065 in. 3
• 19, 363 in.
3
• 9, 682 in. 3
• 16, 747 in. 3
To calculate the volume of the sculpture, we first need to find the volume of the larger hemisphere with a radius of 20 inches, and then subtract the volume of the smaller hemisphere with a radius of 15 inches.
The formula for the volume of a hemisphere is given by V = (2/3) * pi * r^3, where V is the volume, pi is approximately 3.14, and r is the radius.
For the larger hemisphere:
V_large = (2/3) * 3.14 * 20^3
= (2/3) * 3.14 * 8000
≈ 33,493.33 in.^3 (rounded to the nearest hundredth)
For the smaller hemisphere:
V_small = (2/3) * 3.14 * 15^3
= (2/3) * 3.14 * 3375
≈ 14,137.5 in.^3 (rounded to the nearest tenth)
Now, we calculate the volume of the sculpture by subtracting the volume of the smaller hemisphere from the volume of the larger hemisphere:
V_sculpture = V_large - V_small
≈ 33,493.33 - 14,137.5
≈ 19,355.83 in.^3 (rounded to the nearest whole number)
Therefore, the volume of the sculpture is approximately 19,356 in.^3, so the correct answer is:
• 19,356 in.^3