Answer:
See explaination for the details of the answer
Step-by-step explanation:
Defect per opportunity DPO = defects/ no. of opportunities = 2/25 = 0.08
Defect per million opportunities
DPMO = DPO * 1 million
DPMO = 0.08 * 1 million = 80,000
six sigma = 2.9, take the dpmo at higher level if not exact 80800 for 2.9
81 , 243 , 729 the pattern is mutiply each by three
Answer:
•A c-chart is the appropriate control chart
• c' = 8.5
• Control limits, CL = 8.5
Lower control limits, LCL = 0
Upper control limits, UCL = 17.25
Step-by-step explanation:
A c chart is a quality control chart used for the number of flaws per unit.
Given:
Past inspection data:
Number of units= 100
Total flaws = 850
We now have:
c' = 850/100
= 8.5
Where CL = c' = 8.5
For control limits, we have:
CL = c'
UCL = c' + 3√c'
LCL = c' - 3√c'
The CL stands for the normal control limit, while the UCL and LCL are the upper and lower control limits respectively
Calculating the various control limits we have:
CL = c'
CL = 8.5
UCL = 8.5 + 3√8.5
= 17.25
LCL = 8.5 - 3√8.5
= -0.25
A negative LCL tend to be 0. Therefore,
LCL = 0
Let x be the length of the train.
On the basis of the observer;
Speed of the train = x/6
On the basis of the bridge;
Total distance covered by any point of the train= 350+x
Speed = (350+x)/20
Equating the two expressions of speed;
x/6 = (350+x)/20
20(x) = 6(350+x)
20x = 2100+6x
(20-6)x = 2100
14x = 2100
x= 2100/14 = 150 m
Speed = x/6 = (350+x)/20 = 150/6 = 500/20 = 25 m/s.
Therefore,
Length of train = 150 m
Speed of train = 25 m/s
Answer:
81 m/s
Step-by-step explanation:
If the box is sliding at 27 m per second, and it's been sliding for 3 seconds, then you should multiply 3 × 27 and your answer would be 81 m/s