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mylen [45]
3 years ago
7

Name two examples of quadrilaterals,

Mathematics
2 answers:
Nadya [2.5K]3 years ago
5 0

Answer:

Squares, Trapezoids, Rhombuses, Kites, Parellelograms, or Rectangles

Step-by-step explanation:

"Quad" means that there is 4 sides on something, and all of these shapes have 4 sides, so obviously, that means they are quadrilaterals.

kolezko [41]3 years ago
4 0

Answer:

square and rectangle

Step-by-step explanation:

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The second line since it have marks on -1 and on 2
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3 years ago
This exercise illustrates that poor quality can affect schedules and costs. A manufacturing process has 90 customer orders to fi
svp [43]

Answer:

a) 0.0645 = 6.45% probability that the 90 orders can be filled without reordering components.

b) 0.4062 = 40.62%  probability that the 100 orders can be filled without reordering components.

c) 0.9034 = 90.34% probability that the 100 orders can be filled without reordering components

Step-by-step explanation:

For each component, there are only two possible outcomes. Either it is defective, or it is not. The components can be assumed to be independent. This means that we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

3% of the components are identified as defective

This means that p = 0.03

a. If the manufacturer stocks 90 components, what is the probability that the 90 orders can be filled without reordering components?

0 defective in a set of 90, which is P(X = 0) when n = 90. So

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 0) = C_{90,0}.(0.03)^{0}.(0.97)^{90} = 0.0645

0.0645 = 6.45% probability that the 90 orders can be filled without reordering components.

b. If the manufacturer stocks 102 components, what is the probability that the 100 orders can be filled without reordering components?

At most 102 - 100 = 2 defective in a set of 102, so P(X \leq 2) when n = 102

Then

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

In which

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 0) = C_{102,0}.(0.03)^{0}.(0.97)^{102} = 0.0447

P(X = 1) = C_{102,0}.(0.03)^{1}.(0.97)^{101} = 0.1411

P(X = 2) = C_{102,2}.(0.03)^{2}.(0.97)^{100} = 0.2204

P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.0447 + 0.1411 + 0.2204 = 0.4062

0.4062 = 40.62%  probability that the 100 orders can be filled without reordering components.

c. If the manufacturer stocks 105 components, what is the probability that the 100 orders can be filled without reordering components?

At most 105 - 100 = 5 defective in a set of 105, so P(X \leq 5) when n = 105

Then

P(X \leq 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)

In which

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 0) = C_{105,0}.(0.03)^{0}.(0.97)^{105} = 0.0408

P(X = 1) = C_{105,0}.(0.03)^{1}.(0.97)^{104} = 0.1326

P(X = 2) = C_{105,2}.(0.03)^{2}.(0.97)^{103} = 0.2133

P(X = 3) = C_{105,3}.(0.03)^{3}.(0.97)^{102} = 0.2265

P(X = 4) = C_{105,4}.(0.03)^{4}.(0.97)^{101} = 0.1786

P(X = 5) = C_{105,5}.(0.03)^{5}.(0.97)^{100} = 0.1116

P(X \leq 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) = 0.0408 + 0.1326 + 0.2133 + 0.2265 + 0.1786 + 0.1116 = 0.9034

0.9034 = 90.34% probability that the 100 orders can be filled without reordering components

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Step-by-step explanation:

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Step-by-step explanation:

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Answer:

1. x= -12

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Step-by-step explanation:

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2 years ago
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