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Jill has a swimming pool in her backyard. It is shaped like a rectangle and measures approximately 16. 3 feet wide and 26. 2 feet long. It is an average of 5 feet deep. During a few hot weeks during the summer, some water evaporates from the pool, and Jill needs to add 8 inches of water to the depth of the pool, using her garden hose. Although her water pressure varies, the water flows through Jill's garden hose at an average rate of 10 gallons/minute
Convert the dimensions of the pool to meters. Round all meter measurements to one decimal place (nearest tenth of a meter). How many liters will Jill need to add to her pool to return the water level to its original depth? How many gallons of water is this? Reminder: Volume = Length x Width x Depth
How long will Jill need to run the hose? Express your answer in hours and minutes



Answer :

tutorconsortium270

The volume is the product of the area of the top of the pool and the

height of water added which is approximately 8,062 liters.

Response:

  • The volume of water Jill needs to add is 8,062 liters
  • The volume in gallons is 2,129.8 gallons.

How to calculate volume of solids ?

The volume of a solid is used to determine how much space it takes up. The measurement is the number of unit cubes needed to completely fill the solid. The solid has a total of 30 unit cubes, hence its volume is 2 units. 30 cubic units in 3 units, from 5 units.

Calculation

The given dimensions of the pool are;

Width of the pool = 16.3 feet

Length of the pool = 26.2 feet

Height of water to be added = 8 inches

Required:

The volume of water to be added in liters

Solution:

First part

1 feet = 12 inches

8 inches = 8/12 feet = 2/3 feet

Therefore;

Volume of water added is therefore;

Volume, V = Length × Width × Depth

V  = 16.3 x 26.2 x 2/3 = [tex]284\frac{53}{75}[/tex]

Therefore;

The volume added, V = [tex]284\frac{53}{75}[/tex] ft.³

1 ft.³ = 28.31685 L

Therefore;

[tex]284\frac{53}{75} ft^{3}[/tex] = [tex]284\frac{53}{75}[/tex] x 28.31685 ≈ 8062 L

The volume of water that Jill needs to add to her pool is approximately 8062 L.

Second part

1 L = 0.264172 gallons

8062 L = 8062 × 0.264172 gallons ≈ 2129.8 gallons

The volume of water added to the pool in gallons is approximately 2,129.8 gallons

learn more about volume of solids here :

brainly.com/question/21983203

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