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Blababa [14]
2 years ago
5

Which of the following might a cement mason or concrete finisher be responsible for? (Select all that apply).

Engineering
1 answer:
Salsk061 [2.6K]2 years ago
5 0
All of them except the last one. And I’m a concrete mason myself I’m best at finishing
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The most important use of a Pareto diagram is to show the most important factor among the set of factors that have been shown. Along with that it also shows the sources which lead to the common defects in the system and tries to solve those defects which occur most often.

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6. During some actual expansion and compression processes in piston–cylinder devices, the gases have been
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5 0
1 year ago
The no-slip condition is:________
viva [34]

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6 0
3 years ago
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2 years ago
Past evidence shows that when a customer complains of an out-of-orderphone there is an 8% chance that the problem is with the in
Alexxandr [17]

Answer:

a. The expected number of failures due to a problem with inside wiring is 8 failures

b. The probability that at least 10 failures are due to inside wiring is approximately 0.176

c. It will be not unusual

Explanation:

The probability that the problem of an out of order is the inside wiring, P(x) = 8%

The number of complaints in a month period, x = 100

a. The expected number of failures due to a problem with inside wiring, E(x) = x·P(x)

∴ E(x) = 100 × 8% = 8

The expected number of failures due to a problem with inside wiring, E(x) = 8 failures

b. The probability that at least 10, P₁₀, failures are due to inside wiring is given as follows;

The probability of success, P = 0.08, therefore, the probability of failure, q = 1 - 0.08 = 0.92

P = _nC_r·P^r·q^{n-r}

P₀ = ₁₀₀C₀·(0.08)⁰·(0.92)¹⁰⁰ = 0.000239211874657

P₁ = ₁₀₀C₁·(0.08)¹·(0.92)⁹⁹ = 0.00208010325

P₂ = ₁₀₀C₂·(0.08)²·(0.92)⁹⁸ = 0.00895348793

P₃ = ₁₀₀C₃·(0.08)³·(0.92)⁹⁷ = 0.02543309616

P₄ = ₁₀₀C₄·(0.08)⁴·(0.92)⁹⁶ = 0.0536306593

P₅ = ₁₀₀C₅·(0.08)⁵·(0.92)⁹⁵ = 0.0895398833653

P₆ = ₁₀₀C₆·(0.08)⁶·(0.92)⁹⁴ = 0.123279549561

P₇ = ₁₀₀C₃·(0.08)⁷·(0.92)⁹³ = 0.143953759735

P₈ = ₁₀₀C₈·(0.08)⁸·(0.92)⁹² = 0.145518474516

P₉ = ₁₀₀C₉·(0.08)⁹·(0.92)⁹¹ = 0.129349755125

P₁₀ = ₁₀₀C₁₀·(0.08)¹⁰·(0.92)⁹⁰ = 0.10235502362

∴ P₀ + P₁ + P₂ + P₃ + P₄ + P₅ + P₆ + P₇ + P₈ + P₉ + P₁₀ = 0.82433004464

The probability of at least 10 failures are due problem with the inside wiring = 1 - (P₀ + P₁ + P₂ + P₃ + P₄ + P₅ + P₆ + P₇ + P₈ + P₉ + P₁₀) = 1 - 0.82433004464 = 0.175666995536

The probability of at least 10 failures are due problem with the inside wiring = 0.175666995536 ≈ 0.176

c. The probability of at most 5 failures are due problem with the inside wiring = P₀ + P₁ + P₂ + P₃ + P₄ + P₅ = 0.179876441908

Therefore, given that probability of at most 5 failures > The probability of 8 failures it will be not unusual since the cause of failure is more (92%) due to other causes which are more likely and therefore increase in the probability that there are fewer failures due inside wiring

6 0
3 years ago
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