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dmitriy555 [2]
3 years ago
8

HELP!

Engineering
1 answer:
olya-2409 [2.1K]3 years ago
5 0
The thickness is thick
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The number of weaving errors in a twenty-foot by ten-foot roll of carpet has a mean of 0.8 What is the probability of observing
Viktor [21]

Answer:

0.14% probability of observing more than 4 errors in the carpet

Explanation:

When we only have the mean, we use the Poisson distribution.

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}

In which

x is the number of sucesses

e = 2.71828 is the Euler number

\mu is the mean in the given interval.

The number of weaving errors in a twenty-foot by ten-foot roll of carpet has a mean of 0.8.

This means that \mu = 0.8

What is the probability of observing more than 4 errors in the carpet

Either we observe 4 or less errors, or we observe more than 4. The sum of the probabilities of these outcomes is 1. So

P(X \leq 4) + P(X > 4) = 1

We want P(X > 4). Then

P(X > 4) = 1 - P(X \leq 4)

In which

P(X \leq 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)

P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}

P(X = 0) = \frac{e^{-0.8}*(0.8)^{0}}{(0)!} = 0.4493

P(X = 1) = \frac{e^{-0.8}*(0.8)^{1}}{(1)!} = 0.3595

P(X = 2) = \frac{e^{-0.8}*(0.8)^{2}}{(2)!} = 0.1438

P(X = 3) = \frac{e^{-0.8}*(0.8)^{3}}{(3)!} = 0.0383

P(X = 4) = \frac{e^{-0.8}*(0.8)^{4}}{(4)!} = 0.0077

P(X \leq 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 0.4493 + 0.3595 + 0.1438 + 0.0383 + 0.0077 = 0.9986

P(X > 4) = 1 - P(X \leq 4) = 1 - 0.9986 = 0.0014

0.14% probability of observing more than 4 errors in the carpet

5 0
3 years ago
Are ocean currents always cold
swat32

Answer:

The surface ocean currents have a strong effect on Earth's climate. ... However, these areas do not constantly get warmer and warmer, because the ocean currents and winds transport the heat from the lower latitudes near the equator to higher latitudes near the poles.

8 0
3 years ago
Explain the differences between planned and predictive maintenance.
sveticcg [70]

Answer:

Planned maintenance refers to any scheduled activity carried out to check a machine is working ok and diagnose procedures to fix it if need it. On the other hand, predictive mainteance is all the techniques which help to define if a machine requires or not maintenance activities so far.

Explanation:

Planned maintenance is based on preventive routines to ensure a machine is working in acceptable conditions and at the same time prevent them to change to risky values performing acticities like parts replacement, cleaning,  etc. The key of this maintenance is schedule, that is to say, is a maintenance that has to be carried out constantly each certain time. Predictive maintenance is different because it is used to define if a machie needs any kind of inspection or if, on the contrary, the machine can continue operating without any intervention. The good point about predictive maintenance is the capability of telling when a maintenance is required and when is no necessarily required which is ideal to save costs.

7 0
3 years ago
Steam flows steadily through an adiabatic turbine. The inlet conditions of the steam are 10 MPa, 450°C, and 80 m/s, and the exit
8090 [49]

Answer:

a) The change in Kinetic energy, KE = -1.95 kJ

b) Power output, W = 10221.72 kW

c) Turbine inlet area, A_1 = 0.0044 m^2

Explanation:

a) Change in Kinetic Energy

For an adiabatic steady state flow of steam:

KE = \frac{V_2^2 - V_1^2}{2} \\.........(1)

Where Inlet velocity,  V₁ = 80 m/s

Outlet velocity, V₂ = 50 m/s

Substitute these values into equation (1)

KE = \frac{50^2 - 80^2}{2} \\

KE = -1950 m²/s²

To convert this to kJ/kg, divide by 1000

KE = -1950/1000

KE = -1.95 kJ/kg

b) The power output, w

The equation below is used to represent a  steady state flow.

q - w = h_2 - h_1 + KE + g(z_2 - z_1)

For an adiabatic process, the rate of heat transfer, q = 0

z₂ = z₁

The equation thus reduces to :

w = h₁ - h₂ - KE...........(2)

Where Power output, W = \dot{m}w..........(3)

Mass flow rate, \dot{m} = 12 kg/s

To get the specific enthalpy at the inlet, h₁

At P₁ = 10 MPa, T₁ = 450°C,

h₁ = 3242.4 kJ/kg,

Specific volume, v₁ = 0.029782 m³/kg

At P₂ = 10 kPa, h_f = 191.81 kJ/kg, h_{fg} = 2392.1 kJ/kg, x₂ = 0.92

specific enthalpy at the outlet, h₂ = h_1 + x_2 h_{fg}

h₂ = 3242.4 + 0.92(2392.1)

h₂ = 2392.54 kJ/kg

Substitute these values into equation (2)

w = 3242.4 - 2392.54 - (-1.95)

w = 851.81 kJ/kg

To get the power output, put the value of w into equation (3)

W = 12 * 851.81

W = 10221.72 kW

c) The turbine inlet area

A_1V_1 = \dot{m}v_1\\\\A_1 * 80 = 12 * 0.029782\\\\80A_1 = 0.357\\\\A_1 = 0.357/80\\\\A_1 = 0.0044 m^2

3 0
3 years ago
Who plays a role in the financial activities of a company?
KatRina [158]

Hey,

Who plays a role in the financial activities of a company?

<em>O D. Everyone at the company, including managers and employees</em>

3 0
4 years ago
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