Rectangle MNOP with vertices M(-7, 1),
N(4, 1), 0(-4, 4), and P(-7, 4) in the line
y = -x

Rectangle MNOP With Vertices M(-7, 1),N(4, 1), 0(-4, 4), And P(-7, 4) In The Liney = -x

Answers

Answer 1

Answer:

 [tex]M' =(-1,7)[/tex]

[tex]N' =(-1,-4)[/tex]

[tex]O' =(-4,4)[/tex]

[tex]P' =(-4,7)[/tex]

Step-by-step explanation:

Given

[tex]M = (-7, 1)[/tex]

[tex]N = (4, 1)[/tex]

[tex]0 = (-4, 4)[/tex]

[tex]P = (-7, 4)[/tex]

Line: [tex]y = -x[/tex]

Required

The coordinates of M'N'O'P'

The rule for reflection over [tex]y=-x[/tex] is : [tex](x,y) \to (-y,-x)[/tex]

So, we have:

[tex]M = (-7, 1)[/tex]  [tex]\to M' =(-1,7)[/tex]

[tex]N = (4, 1)[/tex] [tex]\to N' =(-1,-4)[/tex]

[tex]0 = (-4, 4)[/tex] [tex]\to O' =(-4,4)[/tex]

[tex]P = (-7, 4)[/tex] [tex]\to P' =(-4,7)[/tex]

So, the points are:

[tex]M' =(-1,7)[/tex]

[tex]N' =(-1,-4)[/tex]

[tex]O' =(-4,4)[/tex]

[tex]P' =(-4,7)[/tex]


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Answer:

sin=The sin of an acute angle is defined in the context of a right triangle.

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Step-by-step explanation:

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Step-by-step explanation:

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Answers

Answer:

0.2773 = 27.73% probability that at the May celebration, exactly two members of the group have May birthdays

Step-by-step explanation:

For each person, there are only two possible outcomes. Either they have a birthday in May, or they do not. The probability of a person having a birthday in May is independent of any other person. This means that the binomial probability distribution is used 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.

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

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

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[tex]p = \frac{31}{365} = 0.0849[/tex]

Group of 20 friends:

This means that [tex]n = 20[/tex]

What is the probability that at the May celebration, exactly two members of the group have May birthdays?

This is P(X = 2).

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

[tex]P(X = 2) = C_{20,2}.(0.0849)^{2}.(0.9151)^{18} = 0.2773[/tex]

0.2773 = 27.73% probability that at the May celebration, exactly two members of the group have May birthdays

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Answers

Answer:

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Step-by-step explanation:

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Answers

Answer:

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Answer:  122 cm

------------------

Explanation:

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==========================================================

Problem 14

Answer: Approximately 57 liters

------------------

Explanation:

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==========================================================

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------------------

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☆*: .。..。.:*☆☆*: .。..。.:*☆☆*: .。..。.:*☆☆*: .。..。.:*☆

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Automated manufacturing operations are quite precise but still vary, often with distribution that are close to Normal. The width in inches of slots cut by a milling machine follows approximately the N(1,0.001)N(1,0.001) distribution. The specifications allow slot widths between 0.999750.99975 and 1.000251.00025. What proportion of slots meet these specifications

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Step-by-step explanation:

The complete question is defined in the attachment file please find it.

Given:

[tex]\mu = 1\\\\\sigma = 0.001 \\\\[/tex]

Converting the Standard Normal:

[tex]\to P(X < x) = P( Z < \frac{( X - \mu )}{\sigma} )\\\\\to P ( 0.9998 < X < 1.0002 ) = P ( Z < \frac{( 1.0002 - 1 )}{0.001}) - P ( Z < \frac{( 0.9998 - 1 )}{0.001})\\\\[/tex]

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Step-by-step explanation:

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Step-by-step explanation:

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Answers

Answer:

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Step-by-step explanation:

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Answers

Answer:

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Step-by-step explanation:

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Answers

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