What are the domain and range of the function represented by the set of
ordered pairs?
{(-6, 12).(-3,-2), (1,0), (14,-4)}

What Are The Domain And Range Of The Function Represented By The Set Ofordered Pairs?{(-6, 12).(-3,-2),

Answers

Answer 1

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Domain = { -6 , -3 , 1 , 14 }

Range = { -4 , -2 , 0 , 12 }

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Thus the correct answer is (( B )) .

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Answer 2

Here, the domain and range of the function represented by the set of ordered pairs {(-6, 12).(-3,-2), (1,0), (14,-4)} is

{-6, -3, 1, 14} and {12, -2, 0, -4} respectively.

What is range?

We know that the range refers to all possible values of y, and the domain refers to all possible values of x.

How to solve it?

For (-6, 12), the x value = -6 and the y value = 12.

For (-3, -2), the x value = -3 and the y value = -2.

For (1, 0), the x value = 1 and the y value = 0.

For (14, -4), the x value = 14 and the y value = -4.

So, the domain is {-6, -3, 1, 14} and the range is {12, -2, 0, -4}.

Therefore the domain and range of the function represented by the set of ordered pairs {(-6, 12).(-3,-2), (1,0), (14,-4)} is

{-6, -3, 1, 14} and {12, -2, 0, -4} respectively. So, option B is correct.

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Answers

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B. If a product were chosen randomly and found to be defective, what is the probability that it was made by machine B3.

Answers

Answer:

a. The probability that the finished product selected is defective is 2.45%.

b. The probability that product chosen randomly was defective and made by machine B3 is 20.41%.

Step-by-step explanation:

Let A represents the defective product.

We also have the following from the question:

P(B1) = Probability or percentage of the made by machine B1 = 30%, or 0.30

P(B2) = Probability or percentage of the made by machine B2 = 45%, or 0,45

P(B3) = Probability or percentage of the made by machine B3 = 25%, or 0.25

P(A/B1) = Probability or percentage of product B1 that is defective = 2%, or 0.02

P(A/B2) = Probability or percentage of product B2 that is defective = 3%, or 0.03

P(A/B3) = Probability or percentage of product B3 that is defective = 2%, or 0.02

We can therefore proceed as follows:

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To determine this, the rules of elimination is applied and we have:

P(A) = (P(B1) * P(A/B1)) + (P(B2) * P(A/B2)) + (P(B3) * P(A/B3)) ………… (1)

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P(A) = 0.0245, or 2.45%

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In southern California, a growing number of individuals pursuing teaching credentials are choosing paid internships over traditional student teaching programs. A group of eight candidates for three local teaching positions consisted of five who had enrolled in paid internships and three who enrolled in traditional student teaching programs. All eight candidates appear to be equally qualified, so three are randomly selected to fill the open positions. Let Y be the number of internship trained candidates who are hired. Find the probability that two or more internship trained candidates are hired.

Answers

Answer:

The required probability = 0.7143

Step-by-step explanation:

From the information given:

From a group of eight candidates

The no. of candidates that enrolled in internships = 5

The no. of candidates that enrolled in teaching = 3

Also, supposed all the eight candidates are equally qualified;

Then, Let assume that:

Y to represent the number of internship trainee candidates hired.

N to represent no. of candidates in a group = 8

r to represent those who enrolled in paid internship = 5

Now, N - r = 3 (for those who enrolled in traditional teaching program)

Suppose; n represent the positions for local teaching which is given as 3;

Then; selecting 3 from 8 whereby some enrolled in internships and some in traditional teaching programs;

Then,  let assume Y is a random variable that follows a hypergeometric distribution; we have:

[tex]p(Y = y) = \left \{ {{\dfrac{ \bigg (^r_y \bigg)\bigg (^{N-r}_{n-y} \bigg) }{ \bigg ( ^N_n \bigg) } } _\atop { ^{0, otherwise} } } \right.[/tex]

[tex]p(Y = y) = \left \{ {{\dfrac{ \bigg (^5_y \bigg)\bigg (^{3}_{3-y} \bigg) }{ \bigg ( ^8_3 \bigg) } } } } \right, y= 0,1,2,3[/tex]

Thus, the probability that two or more internship trained candidates are hired can be computed as:

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[tex]p(Y \geq 2) = \dfrac{ \bigg ( ^5_2\bigg) \bigg ( ^3_1 \bigg)}{\bigg (^8_3 \bigg)} + \dfrac{\bigg (^5_3 \bigg) \bigg (^3_0 \bigg)}{\bigg ( ^8_3\bigg)}[/tex]

[tex]p(Y \geq 2) = \dfrac{40}{56}[/tex]

[tex]\mathbf{p(Y \geq 2) = 0.7143}[/tex]

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