The formula for the volume of a sphere is V = (4/3)(pi)(r^3), where r is the radius of the sphere. Since the diameter of the basketball is 9.2 inches, the radius is half of that, so r = 4.6 inches.
Plugging this into the formula, we get:
V = (4/3)(3.14)(4.6^3)
V = (4/3)(3.14)(97.336)
V = 407.51 inches cubed
Therefore, the volume of the basketball is 407.51 inches cubed.
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) Responses
407.51 in.3 407.51 inches cubed
1,222.54 in.3 1,222.54 in cubed
115.55 in.3 115.55 inches cubed
3,260.11 in.3
7 answers
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) Responses 23.34π yds3 23.34 pi yds cubed 7.2π yds3 7.2 pi yds cubed 24.43π yds3 24.43 pi yds cubed 7.78π yds3
The formula for the volume of a sphere is V = (4/3)(pi)(r^3), where r is the radius of the sphere. Given the radius is 1.8 yards, we can plug it into the formula:
V = (4/3)(pi)(1.8^3)
V = (4/3)(pi)(5.832)
V = 23.34pi yards cubed
Therefore, the volume of the sphere with a radius of 1.8 yards is 23.34pi yards cubed.
V = (4/3)(pi)(1.8^3)
V = (4/3)(pi)(5.832)
V = 23.34pi yards cubed
Therefore, the volume of the sphere with a radius of 1.8 yards is 23.34pi yards cubed.
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) Responses 3,052.1 in.3 3,052.1 inches cubed 190.8 in.3 190.8 inches cubed 381.5 in.3 381.5 inches cubed 113 in.3
The diameter of the basketball is 9 inches, so the radius would be half of that which is 4.5 inches.
Using the formula for the volume of a sphere: V = 4/3 * pi * r^3
Substitute the radius r=4.5 inches into the formula:
V = 4/3 * 3.14 * 4.5^3
V = 4/3 * 3.14 * 91.125
V = 381.6276 inches cubed
Rounding to the nearest tenth, the air needed to fill the basketball is 381.6 inches cubed.
Therefore, the closest answer choice is 381.5 in^3.
Using the formula for the volume of a sphere: V = 4/3 * pi * r^3
Substitute the radius r=4.5 inches into the formula:
V = 4/3 * 3.14 * 4.5^3
V = 4/3 * 3.14 * 91.125
V = 381.6276 inches cubed
Rounding to the nearest tenth, the air needed to fill the basketball is 381.6 inches cubed.
Therefore, the closest answer choice is 381.5 in^3.
The volume of a container measures 4,000 mm3 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) Responses 2,325.3 mm3 2,325.3 mm cubed 1,674.7 mm3 1,674.7 mm cubed 3,900 mm3 3,900 mm cubed 33.5 mm3
First, let's calculate the volume of one marble using the formula for the volume of a sphere: V = (4/3) * pi * r^3, where r is the radius of the marble (2mm).
V = (4/3) * 3.14 * 2^3
V = (4/3) * 3.14 * 8
V = 33.51 mm^3 (rounded to the nearest hundredth)
Next, let's calculate the total volume of the marbles in the box:
Total volume of marbles = 50 marbles * 33.51 mm^3 per marble
Total volume of marbles = 1675.5 mm^3
Now, we need to find the unused space in the box:
Unused space = Total volume of the container - Total volume of the marbles
Unused space = 4000 mm^3 - 1675.5 mm^3
Unused space = 2324.5 mm^3
Rounded to the nearest tenth, the unused space in the box is 2324.5 mm^3.
Therefore, the closest answer choice is 2,325.3 mm^3.
V = (4/3) * 3.14 * 2^3
V = (4/3) * 3.14 * 8
V = 33.51 mm^3 (rounded to the nearest hundredth)
Next, let's calculate the total volume of the marbles in the box:
Total volume of marbles = 50 marbles * 33.51 mm^3 per marble
Total volume of marbles = 1675.5 mm^3
Now, we need to find the unused space in the box:
Unused space = Total volume of the container - Total volume of the marbles
Unused space = 4000 mm^3 - 1675.5 mm^3
Unused space = 2324.5 mm^3
Rounded to the nearest tenth, the unused space in the box is 2324.5 mm^3.
Therefore, the closest answer choice is 2,325.3 mm^3.