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(OFFERING ALL THE POINTS I HAVE) Word Problem. Please help!! Part 1 of problem: The main tank has a radius of 70 feet. What is the volume of the quarter-sphere sized tank? Round your answer to the nearest whole number and use 3.14 for Pi. (Use sphere volume formula) Part 2: The theme park company is building a scale model of the killer whale stadium main show tank for an investor's presentation. Each dimension will be made 6 times smaller to accommodate the mock-up in the presentation room. How many times smaller than the actual volume is the volume of the mock-up? Part 3: Using the information from part 2, answer the following question by filling in the blank: The volume of the actual tank is __% of the mock-up of the tank.

(OFFERING ALL THE POINTS I HAVE) Word Problem. Please help!! Part 1 of problem: The-example-1

2 Answers

4 votes

Answer:

Part 1: 359,007 ft³

Part 2: 216 times smaller

Part 3: 21600%

Step-by-step explanation:

Part 1:

The parameters for the tank are;

The radius of the tank = 70 feet

The volume of a sphere = 4/3·π·r³

Therefore, the volume of a quarter sphere = 1/4×The volume of a sphere

The volume of a quarter sphere = 1/4×4/3·π·r³ = π·r³/3

Plugging in the value for the radius gives

Volume = π×70³/3 = 114,333.33×3.14 = 359,006.7≈ 359,007 ft³.

Part 2:

The dimension of the scale model = 1/6 × Actual dimension

Therefore, we have the radius of the sphere of the scale model = 1/6 × 70

Which gives;

The radius of the sphere of the scale model = 35/3 = 11.67 feet

The volume of the scale model = π·r³/3 = (3.14×11.67³)/3 = 1662.07 ≈ 1662 ft³

The number of times smaller the scale model is than the actual volume = (Actual volume)/(Scale model) = (359,007 ft³)/(1662 ft³) = 216 times

The number of times smaller the scale model is than the actual volume = 216 times = (1/Scale of model)³ = (1/(1/6))³ = 6³.

Part 3:

The percentage of the mock-up, x, to the volume of the actual tank is given as follows

x/100 × 1662 = 359,007

∴ x = 216 × 100 = 21600%

The percentage of the mock-up, to the volume of the actual tank is 21600%.

User Ziollek
by
8.1k points
4 votes

Answer:

Part 1: 359,007 ft³

Part 2: 216 times smaller

Part 3: 21600%

Explanation:

Part 1:

The parameters for the tank are;

The radius of the tank = 70 feet

The volume of a sphere = 4/3·π·r³

Therefore, the volume of a quarter sphere = 1/4×The volume of a sphere

The volume of a quarter sphere = 1/4×4/3·π·r³ = π·r³/3

Plugging in the value for the radius gives

Volume = π×70³/3 = 114,333.33×3.14 = 359,006.7≈ 359,007 ft³.

Part 2:

The dimension of the scale model = 1/6 × Actual dimension

Therefore, we have the radius of the sphere of the scale model = 1/6 × 70

Which gives;

The radius of the sphere of the scale model = 35/3 = 11.67 feet

The volume of the scale model = π·r³/3 = (3.14×11.67³)/3 = 1662.07 ≈ 1662 ft³

The number of times smaller the scale model is than the actual volume = (Actual volume)/(Scale model) = (359,007 ft³)/(1662 ft³) = 216 times

The number of times smaller the scale model is than the actual volume = 216 times = (1/Scale of model)³ = (1/(1/6))³ = 6³.

Part 3:

The percentage of the mock-up, x, to the volume of the actual tank is given as follows

x/100 × 1662 = 359,007

∴ x = 216 × 100 = 21600%

The percentage of the mock-up, to the volume of the actual tank is 21600%.

User Shirish Herwade
by
8.2k points

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