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

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.