Answer:
Answer choice 4. 577 pounds
Step-by-step explanation:
Given a cylindrical steel post with dimensions:
- Radius = half of diameter = 0.5 * 6 = 3 inches [converting to feet, we need to divide inches by 12, so we have [tex]\frac{3}{12}=0.25[/tex] feet ]
Volume of a cylinder is given as:
Volume = [tex]\pi r^{2}h[/tex]
Substituting all the given values we have:
[tex]V=\pi r^{2}h\\V=\pi(0.25)^{2}(6)\\V=0.375\pi\\V=0.375(3.14)\\V=1.1775[/tex] cubic feet
Since, 1 cubic foot of steel weights 490, 1.1775 cubic foot of steel would weigh [tex]490*1.1775=576.97[/tex]
Rounding to nearest pound, we would get [tex]577[/tex] pounds as the answer.
Answer choice 4 is correct.
