Answer:
3 sigma lower control limit = 0.0429
Explanation:
Given.
n = 100
days = 100
Number of defective bulbs = 600 defective bulbs
Let p = Process Average
p = 600/(100*40)
P = 600/4000
P = 0.15
q = 1 - p
q = 1 - 0.15
q = 0.85
3 sigma lower limit = p - 3*√(pq/n)
Using the above formula
Substitute in the values
3 sigma lower control limit = 0.15 - 3 * √(0.15 * 0.85/100)
3 sigma lower control limit= 0.15 - 3√0.001275
3 sigma lower control limit = 0.15 - 3* 0.035707142142714
3 sigma lower control limit = 0.15 - 0.107121426428142
3 sigma lower control limit = 0.04287857357185
3 sigma lower control limit = 0.0429 ---- approximated
Answer:
Yes, there is such a way.
Explanation:
If currents flow in the same direction in two or more long parallel wires, there will be an attractive force between the wires. If the current flows in different directions, there will be a repulsive force between the wires. In this case, these three parallel wires, can be be made to carry current in the same direction, creating an attractive force between all three wires.
Note that it is not possible to have at the least one of them carry current in the opposite direction and still have an attractive current between them.
Answer:
The answer is 12.67 TMU
Explanation:
Recall that,
worker’s eyes travel distance must be = 20 in.
The perpendicular distance from her eyes to the line of travel is =24 in
What is the MTM-1 normal time in TMUs that should be allowed for the eye travel element = ?
Now,
We solve for the given problem.
Eye travel is = 15.2 * T/D
=15.2 * 20 in/24 in
so,
= 12.67 TMU
Therefore, the MTM -1 of normal time that should be allowed for the eye travel element is = 12.67 TMU
Answer:
230 m/s northeast, 1.8 m/s up
Explanation:
204 kilometres = 204000 metres
15.0 minutes = 900 seconds
Velocity = Distance / Time
= 204000 / 900
= 230 m/s northeast (to 2 sf.)
1.6km = 1600 metres
Velocity = 1600 / 900
= 1.8 m/s up (to 2 sf.)
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