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oksian1 [2.3K]
3 years ago
11

Can someone please help I’m very confused

Mathematics
2 answers:
Oxana [17]3 years ago
8 0

y + 6 = 4/3(x - 2)

First, distribute the 4/3

y + 6 = 4/3x - 8/3

Subtract 6 from both sides.

y = 4/3x - 26/3

-26/3 is the y-intercept so start the graph at coordinate (0, -26/3)

After you plot the y-intercept, add 4 to they y-coordinate and 3 to the x-coordinate of the y-intercept to get the next point.

0 + 3 = 3

-26/3 + 4 = -14/3

The next point should be at (3, -14/3)

Add as many points as your professor requires.

yKpoI14uk [10]3 years ago
4 0

Answer:

(0, -26/3)     (3, -14/3)

Step-by-step explanation:

y+6 = 4/3x - 8/3

y = 4/3x - 26/3

(0, -26/3)

-26/3 + 4 = -14/3

 (3, -14/3)

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a consumer magazine counts the number of tissues per box in a random sample of 15 boxes of No- Rasp facial tissues. The sample s
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Answer:

95% confidence interval for the population variance of the number of tissues per box is [5043.11 , 23401.31].

Step-by-step explanation:

We are given that a consumer magazine counts the number of tissues per box in a random sample of 15 boxes of No- Rasp facial tissues. The sample standard deviation of the number of tissues per box is 97.

Firstly, the pivotal quantity for 95% confidence interval for the population variance is given by;

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

where, s^{2}  = sample variance = 97^{2} = 9409

              n = sample of boxes = 15

           \sigma^{2}  = population variance

<em>Here for constructing 95% confidence interval we have used chi-square test statistics.</em>

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

P(5.629 < \chi^{2}__1_4 < 26.12) = 0.95  {As the critical value of chi-square at 14

                                         degree of freedom are 5.629 & 26.12}  

P(5.629 < \frac{(n-1)s^{2} }{\sigma^{2} } < 26.12) = 0.95

P( \frac{5.629 }{(n-1)s^{2} } < \frac{1}{\sigma^{2} } < \frac{26.12 }{(n-1)s^{2} } ) = 0.95

P( \frac{(n-1)s^{2} }{26.12 } < \sigma^{2} < \frac{(n-1)s^{2} }{5.629 } ) = 0.95

<u><em>95% confidence interval for</em></u> \sigma^{2} = [ \frac{(n-1)s^{2} }{26.12 } , \frac{(n-1)s^{2} }{5.629 } ]

                                                  = [ \frac{14 \times 9409 }{26.12 } , \frac{14 \times 9409 }{5.629 } ]

                                                  = [5043.11 , 23401.31]

Therefore, 95% confidence interval for the population variance of the number of tissues per box is [5043.11 , 23401.31].

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