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3 years ago

5

Th e NCAA recommends that a competition diving pool intended for use with two 1-m springboards and two 3-m springboards, in addi

tion to diving platforms set at 5 m, 7.5 m, and 10 m above the water, have a width of 75 ft 1 in., a length of 60 ft, and a minimum water depth of 14 ft 10 in. What is the minimum volume of water such a pool would hold in cubic yards, to the nearest whole number?

1 answer:

Anna [14]3 years ago

6 0

Answer:

The volume of the swimming pool is approximately 2475 yd³ to the nearest whole number

Step-by-step explanation:

The given information are;

The width of the swimming pool = 75 ft. 1 in.

The length of swimming pool = 60 ft.

The depth of the swimming pool = 14 ft. 10 in.

Therefore;

The volume of the swimming pool = Width × Length × Depth

The volume of the swimming pool = 75 ft. 1 in. × 60 ft. × 14 ft. 10 in.

The volume of the swimming pool = 901 in. × 720 in. × 178 in. = 115472160 in.³

1 in.³ = 2.14335 × 10⁻⁵ yd³

∴ 115472160 in.³ = 115472160 in.³ × 2.14335 × 10⁻⁵ yd³/in³ ≈ 2474.969 yd³

Which gives;

The volume of the swimming pool ≈ 2475 yd³ to the nearest whole number.

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Answer:

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Step-by-step explanation:

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3 years ago

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In the illustration below, the three cube-shaped tanks are identical. The spheres in any given tank

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Answer:

1) Volume occupied by the spheres are equal therefore the three tanks contains the same volume of water

2) $Amount \ of \, water \ remaining \ in \, the \ tank \ is \ \frac{x^3(6-\pi) }{6}$

Step-by-step explanation:

1) Here we have;

First tank A

Volume of tank = x³

The volume of the sphere = $\frac{4}{3} \pi r^3$

However, the diameter of the sphere = x therefore;

r = x/2 and the volume of the sphere is thus;

volume of the sphere = $\frac{4}{3} \pi \frac{x^3}{8}$ = $\frac{1}{6} \pi x^3$

For tank B

Volume of tank = x³

The volume of the spheres = $8 \times \frac{4}{3} \pi r^3$

However, the diameter of the spheres 2·D = x therefore;

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volume of the spheres = $8 \times \frac{4}{3} \pi (\frac{x}{4})^3= \frac{x^3 \times \pi }{6}$

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Volume of tank = x³

The volume of the spheres = $64 \times \frac{4}{3} \pi r^3$

However, the diameter of the spheres 4·D = x therefore;

r = x/8 and the volume of the sphere is thus;

volume of the spheres = $64 \times \frac{4}{3} \pi (\frac{x}{8})^3= \frac{x^3 \times \pi }{6}$

Volume occupied by the spheres are equal therefore the three tanks contains the same volume of water

2) For the 4th tank, we have;

number of spheres on side of the tank, n is given thus;

n³ = 512

∴ n = ∛512 = 8

Hence we have;

Volume of tank = x³

The volume of the spheres = $512 \times \frac{4}{3} \pi r^3$

However, the diameter of the spheres 8·D = x therefore;

r = x/16 and the volume of the sphere is thus;

volume of the spheres = $512\times \frac{4}{3} \pi (\frac{x}{16})^3= \frac{x^3 \times \pi }{6}$

Amount of water remaining in the tank is given by the following expression;

Amount of water remaining in the tank = Volume of tank - volume of spheres

Amount of water remaining in the tank = $x^3 - \frac{x^3 \times \pi }{6} = \frac{x^3(6-\pi) }{6}$

$Amount \ of \ water \, remaining \, in \, the \ tank = \frac{x^3(6-\pi) }{6}$ .

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