It would cover 5 sq. ft. one foot deep, or 60 sq. ft. one inch deep.
Volume required = 95 sq ft * 2 inch = 95 sq ft * 1/6 sq ft = 95/6 cu ft = 15.833... cu ft. So number of bags @5 cu ft reqd = 15.833.../5 = 3.166... bags ie 4 bags.
6"by 6"by" 6"= .5 Cu Ft
12 in = 1 ft Volume = 311 sq ft × 2 in = 311 sq ft × (2 ÷ 12) ft = 51 5/6 cu ft (≈ 51.8 cu ft) = 51 5/6 ÷ 0.8 cu ft/bag = 64 19/24 bags ≈ 64.8 bags → you need to buy 65 bags.
As it is a hole, there is no dirt in it.However, assuming you have a cylindrical hole 3 ft wide, the volume of dirt removed to make the hole is:volume = π x radius2 x depth= π x (3/2)2 x 5 cu ft= 45π/4 cu ft≈35.34 cu ftIf your hole is a cuboid, that is it is 3 ft x 3ft by 5 ft deep, then the volume of dirt removed is:volume = length x width x depth= 3 x 3 x 5 cu ft= 45 cu ft.
It would cover 5 sq. ft. one foot deep, or 60 sq. ft. one inch deep.
Volume required = 95 sq ft * 2 inch = 95 sq ft * 1/6 sq ft = 95/6 cu ft = 15.833... cu ft. So number of bags @5 cu ft reqd = 15.833.../5 = 3.166... bags ie 4 bags.
5 cu/yrds x 27 cu/ft = 135 cu/ft
6"by 6"by" 6"= .5 Cu Ft
12 in = 1 ft Volume = 311 sq ft × 2 in = 311 sq ft × (2 ÷ 12) ft = 51 5/6 cu ft (≈ 51.8 cu ft) = 51 5/6 ÷ 0.8 cu ft/bag = 64 19/24 bags ≈ 64.8 bags → you need to buy 65 bags.
82,653,950,016
5
Just type this into Google,(350 *5/12 cu ft) to cu ydand get the answer of5.40123457 cu yd
1 cu. yd of cured concrete is about 3600lbs. 1 cu. ft of cured concrete is about 134 lbs. (rounded up) 4 ft x 4 ft x 5/6 ft = 13 1/3 cu. ft 13 1/3 cu. ft = 1787lbs (rounded up)
As it is a hole, there is no dirt in it.However, assuming you have a cylindrical hole 3 ft wide, the volume of dirt removed to make the hole is:volume = π x radius2 x depth= π x (3/2)2 x 5 cu ft= 45π/4 cu ft≈35.34 cu ftIf your hole is a cuboid, that is it is 3 ft x 3ft by 5 ft deep, then the volume of dirt removed is:volume = length x width x depth= 3 x 3 x 5 cu ft= 45 cu ft.
125
2 ft * 13ft * 5 in = 2 ft * 13 ft * 5/12 ft =2*13*5/12 cubic ft = 10.833... cu ft