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mel-nik [20]
2 years ago
6

The side of a square is Two raised to the seven halves power inches. Using the area formula A = s2, determine the area of the sq

uare.
Mathematics
2 answers:
AlladinOne [14]2 years ago
4 0

Answer:

Step-by-step explanation:

Area of a square = s²

s is the side length of the square

Given

s = 2^{7 1/2}

s = 2^15/2

Area = ( 2^15/2)²

Area = 2^15

Hence two area of the square is 2^15 inches

ololo11 [35]2 years ago
4 0

Answer:

A = 128 square inches

Step-by-step explanation:

A square is a figure with equal length of sides.

Given a square of side, s, 2^{\frac{7}{2} } inches.

Area of a square, A = length of side x length of side

                            = s x s

                            = s^{2}

Thus,

A = s^{2}

   = (2^{\frac{7}{2} } )^{2}

   = 2^{\frac{14}{2} }

   = 2^{7}

   = 128

A = 128 square inches

The are of the square is 128 square inches.

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GreenBeam Ltd. claims that its compact fluorescent bulbs average no more than 3.50 mg of mercury. A sample of 25 bulbs shows a m
Harrizon [31]

Answer:

(a) Null Hypothesis, H_0 : \mu \leq 3.50 mg  

     Alternate Hypothesis, H_A : \mu > 3.50 mg

(b) The value of z test statistics is 2.50.

(c) We conclude that the average mg of mercury in compact fluorescent bulbs is more than 3.50 mg.

(d) The p-value is 0.0062.

Step-by-step explanation:

We are given that Green Beam Ltd. claims that its compact fluorescent bulbs average no more than 3.50 mg of mercury. A sample of 25 bulbs shows a mean of 3.59 mg of mercury. Assuming a known standard deviation of 0.18 mg.

<u><em /></u>

<u><em>Let </em></u>\mu<u><em> = average mg of mercury in compact fluorescent bulbs.</em></u>

So, Null Hypothesis, H_0 : \mu \leq 3.50 mg     {means that the average mg of mercury in compact fluorescent bulbs is no more than 3.50 mg}

Alternate Hypothesis, H_A : \mu > 3.50 mg     {means that the average mg of mercury in compact fluorescent bulbs is more than 3.50 mg}

The test statistics that would be used here <u>One-sample z test</u> <u>statistics</u> as we know about the population standard deviation;

                          T.S. =  \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }  ~ N(0,1)

where, \bar X = sample mean mg of mercury = 3.59

            \sigma = population standard deviation = 0.18 mg

            n = sample of bulbs = 25

So, <em><u>test statistics</u></em>  =  \frac{3.59-3.50}{\frac{0.18}{\sqrt{25} } }

                               =  2.50

The value of z test statistics is 2.50.

<u>Now, P-value of the test statistics is given by;</u>

         P-value = P(Z > 2.50) = 1 - P(Z \leq 2.50)

                                             = 1 - 0.9938 = 0.0062

<em />

<em>Now, at 0.01 significance level the z table gives critical value of 2.3263 for right-tailed test. Since our test statistics is more than the critical values of z, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which </em><em><u>we reject our null hypothesis</u></em><em>.</em>

Therefore, we conclude that the average mg of mercury in compact fluorescent bulbs is more than 3.50 mg.

6 0
3 years ago
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