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jonny [76]
2 years ago
5

Write 0.9 as a fraction

Mathematics
2 answers:
Phoenix [80]2 years ago
6 0

Answer:

9/10

Step-by-step explanation:

hope this helped

Grace [21]2 years ago
5 0

Answer:

The answer is 9/10

Step-by-step explanation:

it's 9/10

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Lisa made a shirt using 1/3 m of blue fabric and 3/5 m of red fabric how many meters of fabric did she use in all
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Answer: 14/15 of a meter

Step-by-step explanation:

5 and 3 LCM is 15.

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The first three steps of completing the square to solve the quadratic equation x^2 +4x-6=0, are shown below
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Answer: x=sqrt10-2, x=-sqrt10-2

Step-by-step explanation: square root both sides of the equation.

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3 years ago
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.

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