Answer:
Mogul executives takes samples of n=8 eagletrons at a time.
They found that the average is 3.25 mph.
For n = 8,
D4 =1.864(table value)
D3=0.136(table value)
The upper and lower control limits are found as
UCL =1.864 ×3.25= 6.058
LCL = 0.136 ×3.25= 0.442
Answer:
I think reduce your following distance
Answer:
At the point when the quantity of bit strings is not exactly the quantity of processors, at that point a portion of the processors would stay inert since the scheduler maps just part strings to processors and not client level strings to processors. At the point when the quantity of part strings is actually equivalent to the quantity of processors, at that point it is conceivable that the entirety of the processors may be used all the while. Be that as it may, when a part string obstructs inside the portion (because of a page flaw or while summoning framework calls), the comparing processor would stay inert. When there are more portion strings than processors, a blocked piece string could be swapped out for another bit string that is prepared to execute, in this way expanding the use of the multiprocessor system.When the quantity of part strings is not exactly the quantity of processors, at that point a portion of the processors would stay inert since the scheduler maps just bit strings to processors and not client level strings to processors. At the point when the quantity of bit strings is actually equivalent to the quantity of processors, at that point it is conceivable that the entirety of the processors may be used at the same time. Be that as it may, when a part string hinders inside the piece (because of a page flaw or while summoning framework calls), the relating processor would stay inert. When there are more portion strings than processors, a blocked piece string could be swapped out for another bit string that is prepared to execute, along these lines expanding the usage of the multiprocessor framework.
Explanation:
voltage, current and resistance are the Volt [ V ], Ampere [ A ] and Ohm [ Ω ]
Answer:
B A and C
Explanation:
Given:
Specimen σ
σ
A +450 -150
B +300 -300
C +500 -200
Solution:
Compute the mean stress
σ
= (σ + σ)/2
σ
= (450 + (-150)) / 2
= (450 - 150) / 2
= 300/2
σ = 150 MPa
σ
= (300 + (-300))/2
= (300 - 300) / 2
= 0/2
σ = 0 MPa
σ
= (500 + (-200))/2
= (500 - 200) / 2
= 300/2
σ = 150 MPa
Compute stress amplitude:
σ
= (σ - σ)/2
σ
= (450 - (-150)) / 2
= (450 + 150) / 2
= 600/2
σ = 300 MPa
σ
= (300- (-300)) / 2
= (300 + 300) / 2
= 600/2
σ = 300 MPa
σ
= (500 - (-200))/2
= (500 + 200) / 2
= 700 / 2
σ = 350 MPa
From the above results it is concluded that the longest fatigue lifetime is of specimen B because it has the minimum mean stress.
Next, the specimen A has the fatigue lifetime which is shorter than B but longer than specimen C.
In the last comes specimen C which has the shortest fatigue lifetime because it has the higher mean stress and highest stress amplitude.
