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ivanzaharov [21]
3 years ago
11

Find the area of a regular hexagon with an apothem 17.3 miles long and a side 20 miles long

Mathematics
1 answer:
Zinaida [17]3 years ago
4 0
The area of a hexagon is \frac{3 \sqrt{3} }{2} a^{2} where a = one side

\frac{3 \sqrt{3} }{2} 20^{2}

\frac{3*200 \sqrt{3} }{2}

\frac{600 \sqrt{3} }{2}

300 \sqrt{3} ≈ 1038 mi²

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natima [27]

This is distributive property and to do distributive property, you distribute what is outside of the parenthesis to the numbers inside the parenthesis.

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3 years ago
The proportion of defective computers built by Byte Computer Corporation is 0.15. In an attempt to lower the defective rate, the
FinnZ [79.3K]

Answer:

a)<em> Null hypothesis : H₀</em>:  the proportion of defective item of computer has been lowered. That is P < 0.15

<u><em>Alternative hypothesis: H₁:</em></u> The proportion of defective item of computer

has been higher. That is P> 0.15 (Right tailed test)

b)    Test statistic   Z = \frac{p-P}{\sqrt{\frac{PQ}{n} } }

c)     Calculate the value of the test statistic = 0.991

d) The critical value at 0.01 level of significance = Z₀.₀₁ = 2.57

e) Null hypothesis accepted at 0.01 level of significance

f) we accepted null hypothesis.

  Hence t<em>he proportion of defective item of computer has been lowered. </em>

Step-by-step explanation:

<u>Step(i)</u>:-

<em>Given the sample size 'n' = 42</em>

Given random sample of 42 computers were tested revealing a total of 4 defective computers.

The defective computers 'x' = 4

<em>The sample proportion of defective computers </em>

                                                                p = \frac{x}{n} = \frac{4}{42} = 0.095

<em>Given The Population proportion 'P' = 0.15</em>

<em>The level of significance ∝=0.01</em>

<u>Step(ii)</u>:-

a)<em> Null hypothesis : H₀</em>:  the proportion of defective item of computer has been lowered. That is P < 0.15

<u><em>Alternative hypothesis: H₁:</em></u> The proportion of defective item of computer

has been higher. That is P> 0.15 (Right tailed test)

b)

    Test statistic   Z = \frac{p-P}{\sqrt{\frac{PQ}{n} } }

                       

c)      

                 Z = \frac{0.095-0.15}{\sqrt{\frac{0.15(0.85)}{42} } }

                 z = \frac{-0.055}{\sqrt{0.00303} } = - 0.9991      

                     

  Calculate the value of the test statistic Z = - 0.9991

                                   |Z| = |- 0.9991| = 0.991

<u>Step(iii)</u>:-

d)

        The critical value at 0.01 level of significance = Z₀.₀₁ = 2.57

e)   Calculate the value of the test statistic Z = 0.991 < 2.57  at 0.01 level of significance.

<u><em>Conclusion</em></u>:-

    Hence the null hypothesis is accepted at 0.01 level of significance.

f)

<em>     The proportion of defective item of computer has been lowered.</em>

 

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