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MatroZZZ [7]
3 years ago
6

Can someone please help me on 3,4, and 5

Mathematics
2 answers:
alexandr1967 [171]3 years ago
8 0
A=45 b=60 c=90 I think
mamaluj [8]3 years ago
3 0

This is algebra and rearranging equations.

Let's start with question 3. The sum of all the angles in the triangle is 180 degrees. Therefore, 45 + 2k + k = 180. We can solve this:

180 - 45 = 3k by rearranging the equation so 135 = 3k, so k = 135/3 = 45 degrees. We can now see that the triangle is isosceles, with two angles being both 45 degrees.

The other two follow the same process:

4: a + 2a + 2a+ a = 360 so 6a = 360

a = 360/6 = 60 degrees

5: b + 3/2 b+ (b+45) + (2b-90) + 90 = 540

5b + 1/2b + 45 = 540 (because the -90 and +90 cancel each other out)

so 5 1/2b = 495

b = 90

If you use each letter and substitute in the values into the angles, you will find that they all add up to the sum of angle measures of each shape.

Hope I helped!

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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
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(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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