Answer
given,
thickness of a flange on an aircraft component is uniformly distributed between 0.95 and 1.05 millimeters.
X = U[0.95,1.05] 0.95≤ x ≤ 1.05
the cumulative distribution function of flange
F(x) = P{X≤ x}=
=
b) P(X>1.02)= 1 - P(X≤1.02)
= 
= 0.3
c) The thickness greater than 0.96 exceeded by 90% of the flanges.
d) mean = 
= 1
variance = 
= 0.000833
I would half 198, get
99 and add a 0, this is how you multiply by 5 in your head
198*5 = 990
990 + 200 = 1190 - 2
= 1188
Alternatively we can round 198 to 200: then multiply by 6
200*6 = 1200
Then We subtract the 6 "2's" that we added
1200 - 12 = 1188
The expression for this is:
(6*200) - (6 * 2)
Answer:
The correct options are 1, 3 and 4.
Step-by-step explanation:
We need to find the expressions whose simplified form is a rational number.
Rational number: If a number is defined in the form of p/q where p and q are integers and q≠0, then it is called a rational number.
For example: 0,2, 4.3 etc.
Irrational number: If a number can not defined in the form of p/q, where p and q are integers and q≠0, then it is called an irrational number.
First expression is

12 is a rational number.
Second expression is

is an irrational number.
Third expression is

21 is a rational number.
Fourth expression is

5 is a rational number.
Therefore, the correct options are 1, 3 and 4.
Answer:
A) 
B) 
C) for n = 2
= 1
for n = 3
= 3
for n = 4
= 8
for n = 5
= 19
Step-by-step explanation:
A) A recurrence relation for the number of bit strings of length n that contain a pair of consecutive Os can be represented below
if a string (n ) ends with 00 for n-2 positions there are a pair of consecutive Os therefore there will be :
strings
therefore for n ≥ 2
The recurrence relation for the number of bit strings of length 'n' that contains consecutive Os
b ) The initial conditions
The initial conditions are : 
C) The number of bit strings of length seven containing two consecutive 0s
here we apply the re occurrence relation and the initial conditions
for n = 2
= 1
for n = 3
= 3
for n = 4
= 8
for n = 5
= 19