Answer: 0.1289
Step-by-step explanation:
Given : The proportion of all students at a large university are absent on Mondays. : 
Sample size : 
Mean : 
Standard deviation = 

Let x be a binomial variable.
Using the standard normal distribution table ,
(1)
Z score fro normal distribution:-

For x=4

For x=3

Then , from (1)
Hence, the probability that four students are absent = 
Solution:

We have to find the remainder when f(x) is divided by 
x²-1=0
x=
So, remainder is 13 and -13.
We have to find the" ratio of the area of sector ABC to the area of sector DBE".
Now,
the general formula for the area of sector is
Area of sector= 1/2 r²θ
where r is the radius and θ is the central angle in radian.
180°= π rad
1° = π/180 rad
For sector ABC, area= 1/2 (2r)²(β°)
= 1/2 *4r²*(π/180 β)
= 2r²(π/180 β)
For sector DBE, area= 1/2 (r)²(3β°)
= 1/2 *r²*3(π/180 β)
= 3/2 r²(π/180 β)
Now ratio,
Area of sector ABC/Area of sector DBE =
= 4/3
Answer:
The probability that no flaws occur in a certain portion of wire of length 5 millimeters = 1.1156 occur / millimeters
Step-by-step explanation:
<u>Step 1</u>:-
Given data A copper wire, it is known that, on the average, 1.5 flaws occur per millimeter.
by Poisson random variable given that λ = 1.5 flaws/millimeter
Poisson distribution 
<u>Step 2:</u>-
The probability that no flaws occur in a certain portion of wire

On simplification we get
P(x=0) = 0.223 flaws occur / millimeters
<u>Conclusion</u>:-
The probability that no flaws occur in a certain portion of wire of length 5 millimeters = 5 X P(X=0) = 5X 0.223 = 1.1156 occur / millimeters