Amanager of a large computer network has developed the following probability distribution of the number of interruptions per day: interruptions (x), p(x). 0, 0.32. 1, 0.35. 2, 0.18. 3, 0.08. 4, 0.04. 5, 0.02. 6, 0.01. compute the expected number of interruptions per day. compute the standard deviation.

Step-by-step explanation:

Given that a manager of a large computer network has developed the following probability distribution of the number of interruptions per day

X denotes interruptions and P(X) probability

 x*p(X)x^2p(x)

00.3200

10.350.350.35

20.180.360.72

30.080.240.72

40.040.160.64

50.020.10.5

60.010.060.36

Total11.273.29

Variance1.6771  

std dev1.295028957  

Expected value = sum of all products of x with corresponding p

= 1.27

Answer is 1.27

Std dev = sqrt of variance = 1.2950

stop cheating

step-by-step explanation:

answer:   28.56 feet (approx)

step-by-step explanation:

since, triangle abc has measure of angle c equal to 55 degrees, measure of angle abc equal to 90 degrees, and length of bc equal to 20 feet.

we have to find out the measurement of segment ab.

therefore, tan 55° =

but bc = 20 feet.

⇒ tan 55 ° =  

⇒ ab = 20 × tan 55 °

⇒ ab = 20 × 1.42814800674

⇒ ab = 28.5629601348   ≈ 28.56 feet