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Andrew [12]
3 years ago
15

A line of the equation has a slope of a/b, where a and b are integers.

Mathematics
1 answer:
ElenaW [278]3 years ago
7 0

Answer:

-b/a

Step-by-step explanation:

The slopes of perpendicular lines are negative reciprocals. That means that their product is -1.

slope of line = a/b

slope of perpendicular = -b/a

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5/8x + 10 for x = -8
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\mathrm{Remove\:parentheses}:\quad \left(-a\right)=-a\\=-\frac{5}{8}\cdot \:8+10\\\frac{5}{8}\cdot \:8=5\\=-5+10\\\mathrm{Add/Subtract\:the\:numbers:}\:-5+10=5\\=5

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4 years ago
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Miguel took 7 hours to read a 259-page book and Julissa took 5 hours to read a 175-page book. Who read more pages per hour?
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M: 259/7=37 pages per hour
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3 years ago
The weights of 83 randomly selected windshields were found to have a variance of 1.88. Construct the 95% confidence interval for
Morgarella [4.7K]

Answer:

95% confidence interval for the population variance = (1.42 , 2.62).

Step-by-step explanation:

We are given that the weights of 83 randomly selected windshields were found to have a variance of 1.88.

<em>So, firstly the pivotal quantity for 95% confidence interval for the population variance is given by;</em>

        P.Q. = \frac{(n-1)s^{2} }{\sigma^{2} } ~ \chi^{2} __n_-_1

where, s^{2} = sample variance = 1.88

           \sigma^{2} = population variance

            n = sample of windshields = 83

So, 95% confidence interval for population variance, \sigma^{2} is;

P(58.85 < \chi^{2} __8_2 < 108.9) = 0.95 {As the table of \chi^{2} at 82 degree of freedom

                                              gives critical values of 58.85 & 108.9}

P(58.85 < \frac{(n-1)s^{2} }{\sigma^{2} } < 108.9) = 0.95

P( \frac{ 58.85}{(n-1)s^{2} } < \frac{1 }{\sigma^{2} } < \frac{108.9}{(n-1)s^{2} } ) = 0.95

P( \frac{ (n-1)s^{2}}{108.9 } < \sigma^{2} < \frac{ (n-1)s^{2}}{58.85 } ) = 0.95

<em><u>95% confidence interval for</u></em> \sigma^{2} = ( \frac{ (n-1)s^{2}}{108.9 } , \frac{ (n-1)s^{2}}{58.85 } )

                                                  = ( \frac{ (83-1)\times 1.88}{108.9 } , \frac{ (83-1)\times 1.88}{58.85 } )

                                                  = (1.42 , 2.62)

Therefore, 95% confidence interval for the population variance of the weights of all windshields in this factory is (1.42 , 2.62).

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