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a_sh-v [17]
3 years ago
6

What is the slope of a line that is parallel to a line with slope of m=-6/5.What is the slope of the line that is perpendicular

to a line with a slope of m=6/5.Explain how you know
Mathematics
1 answer:
Verdich [7]3 years ago
8 0

Answer:

a) m = -6/5

b) m = -5/6

Step-by-step explanation:

The slopes of parallel lines are the same. The parallel line will have a slope equal to that of the line it is parallel to, -6/5.

__

The slopes of perpendicular lines are the opposite of the reciprocal of one another. The perpendicular line will have a slope that is the negative reciprocal of the slope of the one it is perpendicular to: -1/m = -1/(6/5) = -5/6.

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Answer:

68%

Explanation:

17/25 = 0.68

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Which tree diagram shows all of the possible outcomes for tossing a coin and rolling a fair number pyramid that
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D : This is because there are 8 outcomes and 2 events.

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Step-by-step explanation:

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3 years ago
A university dean is interested in determining the proportion of students who receive some sort of financial aid. Rather than ex
Olenka [21]

Answer:

a

 The  90% confidence interval that  estimate the true proportion of students who receive financial aid is

     0.533  <  p <  0.64

b

   n = 1789

Step-by-step explanation:

Considering question a

From the question we are told that

      The sample size is  n = 200

      The number of student that receives financial aid is k = 118

Generally the sample proportion is  

      \^ p = \frac{k}{n}

=>   \^ p = \frac{118}{200}

=>   \^ p = 0.59

From the question we are told the confidence level is  90% , hence the level of significance is    

      \alpha = (100 - 90 ) \%

=>   \alpha = 0.10

Generally from the normal distribution table the critical value  of \frac{\alpha }{2}  is  

   Z_{\frac{\alpha }{2} } =  1.645

Generally the margin of error is mathematically represented as  

     E =  Z_{\frac{\alpha }{2} } * \sqrt{\frac{\^ p (1- \^ p)}{n} }

 =>E =  1.645 * \sqrt{\frac{0.59 (1- 0.59)}{200} }

=>  E = 0.057

Generally 90% confidence interval is mathematically represented as  

      \^ p -E <  p <  \^ p +E

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Considering question b

From the question we are told that

    The margin of error  is  E = 0.03

From the question we are told the confidence level is  99% , hence the level of significance is    

      \alpha = (100 - 99 ) \%

=>   \alpha = 0.01

Generally from the normal distribution table the critical value  of   is  

   Z_{\frac{\alpha }{2} } = 2.58

Generally the sample size is mathematically represented as      

        [\frac{Z_{\frac{\alpha }{2} }}{E} ]^2 * \^ p (1 - \^ p )

=>      n = [\frac{2.58}{0.03} ]^2 * 0.59 (1 - 0.59 )

=>      n = 1789

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Answer:

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