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SashulF [63]
2 years ago
10

When a process is in a state of statistical control, all of the points on a control chart should fall within the control limits.

However, it is undesirable that all of the points should fall extremely near, or exactly on, the centerline of the control chart. Why?
Engineering
1 answer:
Mars2501 [29]2 years ago
4 0

it is undesirable that all of the points should fall extremely near, or exactly on, the centerline of the control chart Because:

  • It can make it difficult  to interpret patterns and draw conclusions in a run chart.
  • It can lead hinder the process, and make it unstable.
<h3>When a data point falls outside the control limits of a run chart ?</h3>

If a data point falls is said to outside the control limits, we can say that the process is said to be out of one's control and it is good that an investigation is done to ascertain and remove the cause or causes.

So, it is undesirable that all of the points should fall extremely near, or exactly on, the centerline of the control chart Because:

  • It can make it difficult  to interpret patterns and draw conclusions in a run chart.
  • It can lead hinder the process, and make it unstable.

Learn more about data point from

brainly.com/question/3514929

#SPJ1

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How can statistical analysis of a dataset inform a design process
Shtirlitz [24]

Answer:

Explanation:.

3 0
3 years ago
Determine (a) the principal stresses and (b) the maximum in-plane shear stress and average normal stress at the point. Specify t
raketka [301]

Answer:

a) 53 MPa,  14.87 degree

b) 60.5 MPa  

Average shear = -7.5 MPa

Explanation:

Given

A = 45

B = -60

C = 30

a) stress P1 = (A+B)/2 + Sqrt ({(A-B)/2}^2 + C)

Substituting the given values, we get -

P1 = (45-60)/2 + Sqrt ({(45-(-60))/2}^2 + 30)

P1 = 53 MPa

Likewise P2 = (A+B)/2 - Sqrt ({(A-B)/2}^2 + C)

Substituting the given values, we get -

P1 = (45-60)/2 - Sqrt ({(45-(-60))/2}^2 + 30)

P1 = -68 MPa

Tan 2a = C/{(A-B)/2}

Tan 2a = 30/(45+60)/2

a = 14.87 degree

Principal stress

p1 = (45+60)/2 + (45-60)/2 cos 2a + 30 sin2a = 53 MPa

b) Shear stress in plane

Sqrt ({(45-(-60))/2}^2 + 30) = 60.5 MPa

Average = (45-(-60))/2 = -7.5 MPa

5 0
3 years ago
░░░░░░░░░░░░░░░░░░░▄▀▐░░░▌
allsm [11]

Answer:

wow

Explanation:

that's so cute!!!!!!!!

3 0
3 years ago
The 1000-lb elevator is hoisted by the pulley system and motor M. The motor exerts a constant force of 500 lb on the cable. The
klemol [59]

The power that must be supplied to the motor is 136 hp

<u>Explanation:</u>

Given-

weight of the elevator, m = 1000 lb

Force on the table, F = 500 lb

Distance, s = 27 ft

Efficiency, ε = 0.65

Power  = ?

According to the equation of motion:

F = ma

3(500) - 1000 = \frac{1000}{32.2} * a

a = 16.1 ft/s²

We know,

v^2 - u^2 = 2a (S - So)\\\\v^2 - (0)^2 = 2 * 16.1 (27-0)\\\\v = 29.48m/s

To calculate the output power:

Pout = F. v

Pout = 3 (500) * 29.48

Pout = 44220 lb.ft/s

As efficiency is given and output power is known, we can calculate the input power.

ε = Pout / Pin

0.65 = 44220 / Pin

Pin = 68030.8 lb.ft/s

Pin = 68030.8 / 500 hp

     = 136 hp

Therefore, the power that must be supplied to the motor is 136 hp

5 0
4 years ago
Hi, any kind of help on these questions will be appreciated.
Zielflug [23.3K]

Answer:

IDK

Explanation:

8 0
3 years ago
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