Answer:551.3212cm³
Step-by-step explanation:
Find the image attached
The volume is made up of a cone, cylinder and a hemisphere
Volume of the shape = Volume of cone + volume of cylinder + volume of hemisphere
Get the volume of the cone;
Volume of a cone Vc = 1/3πr²h
r is the base radius = 3.5cm
Height = 10cm
Vc = 1/3π(3.5)²(6)
Vc = 1/3π(12.25)(6)
Vc = 12.25 * 2π
Vc = 24.5π cm³
Get the volume of the cylinder;
Vcy = πr²h
r = 3.5cm
h = 10cm
Vcy = π(3.5)²(10)
Vcy = π(12.25)(10)
Vcy = π(122.5)
Vcy = 122.5π cm³
Get rhe volume of the hemisphere;
Volume of hemisphere = 2/3 πr³
r = 3.5cm
Vh = 2/3 π(3.5)³
Vh = 2/3π(42.875)
Vh = 28.58π cm³
Volume of the shape = VC + Vcy + Vh
Volume of the shape = 24.5π+122.5π+28.58π
Volume of the shape = 175.58π
<em>Volume of the shape = 551.3212cm³</em>
The answer would be D i think because if the two angles are to equal 72, and one of the angles equals 42, 72-42= 30. to find x, you’d do 30/6 which equals 5. so x=5 and the angle measure would be 30
The expected length of code for one encoded symbol is

where
is the probability of picking the letter
, and
is the length of code needed to encode
.
is given to us, and we have

so that we expect a contribution of

bits to the code per encoded letter. For a string of length
, we would then expect
.
By definition of variance, we have
![\mathrm{Var}[L]=E\left[(L-E[L])^2\right]=E[L^2]-E[L]^2](https://tex.z-dn.net/?f=%5Cmathrm%7BVar%7D%5BL%5D%3DE%5Cleft%5B%28L-E%5BL%5D%29%5E2%5Cright%5D%3DE%5BL%5E2%5D-E%5BL%5D%5E2)
For a string consisting of one letter, we have

so that the variance for the length such a string is

"squared" bits per encoded letter. For a string of length
, we would get
.