if it has a diameter of 8, that means its radius is half that, or 4.
![\bf \textit{volume of a cone}\\\\ V=\cfrac{\pi r^2 h}{3}~~ \begin{cases} r=radius\\ h=height\\[-0.5em] \hrulefill\\ r=4\\ h=5 \end{cases}\implies V=\cfrac{\pi (4)^2(5)}{3}\implies V=\cfrac{80\pi }{3} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill \stackrel{using~\pi =3.14}{V= 83.7\overline{3}}~\hfill](https://tex.z-dn.net/?f=%20%5Cbf%20%5Ctextit%7Bvolume%20of%20a%20cone%7D%5C%5C%5C%5C%0AV%3D%5Ccfrac%7B%5Cpi%20r%5E2%20h%7D%7B3%7D~~%0A%5Cbegin%7Bcases%7D%0Ar%3Dradius%5C%5C%0Ah%3Dheight%5C%5C%5B-0.5em%5D%0A%5Chrulefill%5C%5C%0Ar%3D4%5C%5C%0Ah%3D5%0A%5Cend%7Bcases%7D%5Cimplies%20V%3D%5Ccfrac%7B%5Cpi%20%284%29%5E2%285%29%7D%7B3%7D%5Cimplies%20V%3D%5Ccfrac%7B80%5Cpi%20%7D%7B3%7D%0A%5C%5C%5C%5C%5B-0.35em%5D%0A%5Crule%7B34em%7D%7B0.25pt%7D%5C%5C%5C%5C%0A~%5Chfill%20%5Cstackrel%7Busing~%5Cpi%20%3D3.14%7D%7BV%3D%2083.7%5Coverline%7B3%7D%7D~%5Chfill%20)
Answer and Step-by-step explanation:
Considering the table attached.
(a) over 9.5 kg;
μ = 8
σ = 0.9
z = 9.5 - 8/0.9 ≈ 1.67
P (Z > 1.67) = 0.5 - P(0<Z<1.67) = 0.5 - 0.4525 = 0.0475
(b) at most 8.6 kg;
z = 8.6-8/0.9 ≈ 0.67
P(Z < 0.67) = 0.5 + P(0<Z<0.67) = 0.5 + 0.2486 = 0.7486
(c) between 7.3 and 9.1 kg.
z₁ = 7.3-8/0.9 ≈ -0.78
z₂ = 9.1 - 8/0.9 ≈ 1.22
P(-0.78 < Z < 1.22) = P(0 < Z < 0.78) + P(0 < Z < 1.22) = 0.2823 + 0.3888 = 0.6711
B and C are correct, A and D are incorrect
Answer:
The total area is 3016 square inches, approximately