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ira [324]
3 years ago
6

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

more than 4 errors in the carpet
Engineering
1 answer:
Viktor [21]3 years ago
5 0

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

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simple Brayton cycle using air as the working fluid has a pressure ratio of 10. The minimum and maximum temperatures in the cycl
Irina18 [472]

Answer:

a) 764.45K

b) 210.48 kJ/kg

c) 30.14%

Explanation:

pressure ratio = 10

minimum temperature = 295 k

maximum temperature = 1240 k

isentropic efficiency for compressor = 83%

Isentropic efficiency for turbine = 87%

<u>a) Air temperature at turbine exit </u>

we can achieve this by interpolating for enthalpy

h4 = 783.05 kJ/kg ( calculated in the background ) at state 4 using Table A-17  for  Ideal gas properties of air

T4 ( temperature at Turbine exit ) = 760 + ( 780 - 760 ) (\frac{783.05-778.18}{800.13-778.18} ) = 764.45K

<u>b) The net work output </u>

first we determine the actual work input to compressor

Wc = h2 - h1  ( calculated values )

     = 626.57 - 295.17 =  331.4 kJ/kg

next determine the actual work done by Turbine

Wt = h3 - h4  ( calculated values )

     = 1324.93 - 783.05 = 541.88 kJ/kg

finally determine the network output of the cycle

Wnet = Wt - Wc

         = 541.88 - 331.4  = 210.48 kJ/kg

<u>c) determine thermal efficiency </u>

лth = Wnet / qin  ------ ( 1 )

where ; qin = h3 - h2

<em>equation 1 becomes </em>

лth = Wnet / ( h3 - h2 )

      = 210.48 / ( 1324.93 - 626.57 )

      = 0.3014  =  30.14%

6 0
3 years ago
A hydraulic cylinder is to be used to move a workpiece in a manufacturing operation through a distance of 50 mm in 10 s. A force
svetlana [45]

Answer:

The answer to this question is 1273885.3 ∅

Explanation:

<em>The first step is to determine the required  hydraulic flow rate liquid if working pressure and  if a cylinder with a piston diameter of 100 mm is available.</em>

<em>Given that,</em>

<em>The distance = 50mm</em>

<em>The time t =10 seconds</em>

<em>The force F = 10kN</em>

<em>The piston diameter is = 100mm</em>

<em>The pressure = F/A</em>

<em> 10 * 10^3/Δ/Δ </em>

<em> P = 1273885.3503 pa</em>

<em>Then</em>

<em>Power = work/time  = Force * distance /time</em>

<em> = 10 * 1000 * 0.050/10</em>

<em>which is  =50 watt</em>

<em>Power =∅ΔP</em>

<em>50 = 1273885.3 ∅</em>

5 0
3 years ago
Mahamad Siddiqui sent false emails and letters of recommendation on behalf of individuals without their permission to nominate h
shusha [124]

Mahamad Siddiqui sent false emails and letters of recommendation on behalf of individuals without their permission to nominate himself for the Waterman Award at the National Science Foundation. His earlier emails were offered where he had solicited letters were offered as evidence. Siddiqui claimed that content of earlier emails was hearsay. Do the earlier emails come in is given below

Explanation:

1.Mohamed Siddiqui appeals his convictions for fraud and false statements to a federal agency, and obstruction in connection with a federal investigation.   Siddiqui challenges the district court's admission into evidence of e-mail and foreign depositions.

2.On February 18, 1997, Jodi Saltzman, a special agent with the NSF interviewed Siddiqui at Siddiqui's office at the University of South Alabama.   During the interview, Siddiqui signed a statement admitting that he had nominated himself for the Waterman Award, but that he had permission from Yamada and von Gunten to submit forms on their behalf.   Siddiqui also acknowledged in the statement that Westrick had recommended Siddiqui for a different award, the PECASE Award, but that Siddiqui had changed the wording of the letter to apply to the Waterman Award.   Siddiqui was indicted on April 29, 1997.

3.Siddiqui opposed the taking of the depositions on the grounds that the witnesses' personal presence at trial was necessary, and that Indian travel restrictions for its citizens residing abroad prevented him from traveling to Japan and Switzerland.   Specifically, Siddiqui asserted that because of religious persecution in India his travel to Japan or Switzerland related to the criminal action would put his family members still living in India at risk.   The magistrate judge ruled that the government had carried its burden of showing that Yamada and von Gunten would be unavailable to appear at trial, and instructed that Siddiqui's fear of obtaining a travel visa from India because of the threat of persecution of family members should not preclude the taking of the foreign depositions.

4.Yamada's deposition was taken in Japan on March 6, 1998.   At government expense, Siddiqui's counsel attended the deposition and cross-examined the witness, but was not in telephonic contact with Siddiqui during the deposition.   Yamada testified that on February 1, 1997, she received an e-mail stating that if she received a phone call from the NSF to “please tell good words about me.”   Yamada testified that she knew the e-mail was from Siddiqui because the name on the e-mail had Siddiqui's sender address, and it ended with the name “Mo” which Siddiqui had previously told her was his nickname, and which he had used in previous e-mail.

5.Yamada later admitted to Saltzman that she had not given Siddiqui permission to sign, but had made the earlier representation because she thought Siddiqui would go to jail.

6.During cross-examination of Yamada at the deposition, Siddiqui's counsel introduced an e-mail from Yamada to Siddiqui.   This e-mail contained the same e-mail address for Siddiqui as the e-mail received by Yamada and von Gunten apparently from Siddiqui.

7.Von Gunten's video deposition was taken in Switzerland.   At government expense, Siddiqui's counsel attended the deposition and cross-examined von Gunten.   During the deposition, Siddiqui was in communication with his counsel by telephone.   Von Gunten testified at the deposition that he had not submitted a letter of recommendation in favor of Siddiqui for the Waterman Award, and that he had not given Siddiqui permission to submit such a letter in his name.

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4. A banking system provides users with several services:
VLD [36.1K]

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<h3>What is system Case diagram?</h3>

A use case diagram is known to be a kind of graphical illustration of a users in terms of their various possible association or interactions within any given system.

A use case diagram in banking can be used to prepare, depict and also to know all the functional requirements of the banking system.

Therefore, Give the use case specification for the banking system services and paying a bill online is given in the image attached.

Learn  more about Case diagram from

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