first off let's notice that the height is 11 meters and the volume of the cone is 103.62 cubic centimeters, so let's first convert the height to the corresponding unit for the volume, well 1 meters is 100 cm, so 11 m is 1100 cm.
![\textit{volume of a cone}\\\\ V=\cfrac{\pi r^2 h}{3}~~ \begin{cases} r=radius\\ h=height\\[-0.5em] \hrulefill\\ V=\stackrel{cm^3}{103.62}\\ h=\stackrel{cm}{1100} \end{cases}\implies 103.62=\cfrac{\pi r^2 (1100)}{3} \\\\\\ 3(103.62)=1100\pi r^2\implies \cfrac{3(103.62)}{1100\pi }=r^2 \\\\\\ \sqrt{\cfrac{3(103.62)}{1100\pi }}=r\implies \stackrel{cm}{0.00510199305952} \approx r](https://tex.z-dn.net/?f=%5Ctextit%7Bvolume%20of%20a%20cone%7D%5C%5C%5C%5C%20V%3D%5Ccfrac%7B%5Cpi%20r%5E2%20h%7D%7B3%7D~~%20%5Cbegin%7Bcases%7D%20r%3Dradius%5C%5C%20h%3Dheight%5C%5C%5B-0.5em%5D%20%5Chrulefill%5C%5C%20V%3D%5Cstackrel%7Bcm%5E3%7D%7B103.62%7D%5C%5C%20h%3D%5Cstackrel%7Bcm%7D%7B1100%7D%20%5Cend%7Bcases%7D%5Cimplies%20103.62%3D%5Ccfrac%7B%5Cpi%20r%5E2%20%281100%29%7D%7B3%7D%20%5C%5C%5C%5C%5C%5C%203%28103.62%29%3D1100%5Cpi%20r%5E2%5Cimplies%20%5Ccfrac%7B3%28103.62%29%7D%7B1100%5Cpi%20%7D%3Dr%5E2%20%5C%5C%5C%5C%5C%5C%20%5Csqrt%7B%5Ccfrac%7B3%28103.62%29%7D%7B1100%5Cpi%20%7D%7D%3Dr%5Cimplies%20%5Cstackrel%7Bcm%7D%7B0.00510199305952%7D%20%5Capprox%20r)
Answer:
I think 2. is A i dont know the other two
Step-by-step explanation:
Answer:
0.9
Step-by-step explanation:
Since A and B are independent hence;
P(A∩B) = P(A)P(B)
Given the following
P(A∩B) = 0.27
P(A) = 0.3
Required
P(B)
Substitute into the formula;
P(B) = P(A∩B)/P(A)
P(B) = 0.27/0.3
P(B) = 0.9
Hence the value of P(B) is 0.9
Two-hundred and forty-nine is less than two hundred and seventy-one
Answer:
m=4 , m=-2
Step-by-step explanation:
