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3 years ago

HELPPPP

Mathematics

2 answers:

Nimfa-mama [501]3 years ago

5 0

Answer B. Draw a perpendicular bisector

Step-by-step explanation:

finlep [7]3 years ago

3 0

The answer is B. Draw a perpendicular bisector

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A paper company ships one 110.4 tons of paper in four trucks if Eachtruck carries the same weight how many tons of

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Divide total weight by number of trucks:

110.4 / 4 = 27.6 tons per truck

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4 years ago

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if the length of the rectangle is twice the width and the perimeter of the rectangle is 30 centimeters what is the length of the

Dominik [7]

If the perimeter equals 30 and the length is twice the width, then the length is equal to 20 and the width is equal to 10.

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4 years ago

Joe left the park to go home at 5:00 pm. He was traveling east at 15 mph. Two hours later, Mark left the park heading east to tr

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4 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.

So, 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 = 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

95% confidence interval for $\sigma^{2}$ = ( $\frac{ (n-1)s^{2}}{108.9 }$ , $\frac{ (n-1)s^{2}}{58.85 }$ )

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

= (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).

8 0

3 years ago

Quadrilateral ABCD has vertices at A(-2, -4), B(1, -1), C(-2, 2) and D(-5, -1). What is the name of the figure?

dimaraw [331]

Answer:

a square

Step-by-step explanation:

bc if u plot them on a graph, you get a rhombus but when you turn the page, you also get a square. it said under the rhombus, not a square, so we cross that one out and we hv the res. all the sides are equal so the only answer left is a square

hope it helped xx

4 0

3 years ago