The height of a door will be around 6 feet and 8 inches, so about 7 feet. You will need to convert this to meters. To do this, you should first know that 1 meter = 3.28 feet. Now, divide 7 by 3.28:
7/3.28 = 2.134
So, a reasonable estimate would be 2 meters.
118,813,760 different combination license plates can the country produce
<h3 /><h3>permutations and combinations:</h3>
permutations and combinations, the various ways in which objects from a set may be selected, generally without replacement, to form subsets. This selection of subsets is called a permutation when the order of selection is a factor, a combination when order is not a factor. By considering the ratio of the number of desired subsets to the number of all possible subsets for many games of chance in the 17th century, the French mathematicians Blaise Pascal and Pierre de Fermat gave impetus to the development of combinatorics and probability theory.
given that
in a certain country license plates consist of zero or one digit followed by four or five uppercase letters from the roman alphabet
X = 10 × 26 × 26 × 26 × 26 × 26
= 10 × 11881376
= 118,813,760
118,813,760 different combinations license plates can the country produce
To learn more about combinations:
brainly.com/question/19692242
#SPJ4
Answer:
Height of the silo = 18 feet.
Step-by-step explanation:
From the figure attached BC is the length of the silo and the height of the farmer is 5 ft.
Farmer is standing at 8 ft distance from the silo.
From triangle AEC,
tan(∠CAE) = 
= 
m(∠CAE) = 
= 32°
m∠BAE = 90° - 32° = 58°
From the triangle ABE,
tan58° = 
BE = 8tan58°
BE = 12.8 ft
Total height of the silo = BE + EC
= 12.8 + 5
= 17.8
≈ 18 ft
Therefore, total height of the silo is 18 ft.
Answer:
-207xy
Step-by-step explanation:
-6x^3 - 9xy + 18x
-216x -9xy +18x
-225xy + 18x
-207xy
If I'm not mistaken.
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 ![E[L]=1.375n](https://tex.z-dn.net/?f=E%5BL%5D%3D1.375n)
.
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 ![\mathrm{Var}[L]=1.859n](https://tex.z-dn.net/?f=%5Cmathrm%7BVar%7D%5BL%5D%3D1.859n)
.
