27 divided by 15 is 1.8, so that should be the correct answer.
Answer:
32 in²
Step-by-step explanation:
A= ((a+b)/2)*h
A= ((6+10)/2)*4
A=32
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)
.
Answer:
Step-by-step explanation:
From the information given,
Number of personnel sampled, n = 85
Mean or average = 6.5
Standard deviation of the sample = 1.7
We want to determine the confidence interval for the mean number of years that personnel spent in a particular job before being promoted.
For a 95% confidence interval, the confidence level is 1.96. This is the z value and it is determined from the normal distribution table. We will apply the following formula to determine the confidence interval.
z×standard deviation/√n
= 1.96 × 6.5/√85
= 1.38
The confidence interval for the mean number of years spent before promotion is
The lower end of the interval is 6.5 - 1.38 = 5.12 years
The upper end is 6.5 + 1.38 = 7.88 years
Therefore, with 95% confidence interval, the mean number of years spent before being promoted is between 5.12 years and 7.88 years
Answer:
24 in
Step-by-step explanation:
The diameter of a circle is twice its radius which is 12 * 2 = 24.
