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denis23 [38]
3 years ago
8

Help I will mark you brainly​

Mathematics
2 answers:
algol [13]3 years ago
7 0

Answer:

x = 20

y = 10√3 or 17.3

Step-by-step explanation:

This is a special 30° 60° 90° right triangle

In this special triangle if the side length that sees 30° is represented by x and its given as 10 the side length that sees 90° would be twice of it so it's represented by 2x = 20 and the side length that sees 60° is x√3 = 10√3

Ludmilka [50]3 years ago
5 0

Answer:

<em>Hi</em><em> </em><em>there</em><em>!</em><em>!</em>

<em>The</em><em> </em><em>answer</em><em> </em><em>would be</em><em> </em><em>x</em><em>=</em><em>2</em><em>0</em><em> </em><em>and</em><em> </em><em>y</em><em>=</em><em> </em><em>1</em><em>0</em><em> </em><em>root</em><em> </em><em> </em><em>3</em><em> </em><em>or</em><em> </em><em>on</em><em> </em><em>decimal</em><em> </em><em>it's</em><em> </em><em>1</em><em>7</em><em>.</em><em>3</em><em>2</em><em>.</em>

<em>explanation</em><em> </em><em>look</em><em> </em><em>in</em><em> </em><em>picture</em><em>,</em><em> </em><em>alright</em><em>. </em>

<em><u>I</u></em><em><u> </u></em><em><u>hope</u></em><em><u> </u></em><em><u>it will</u></em><em><u> </u></em><em><u>help u</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em>

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4 0
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a) <em> standard error of the mean =10.06</em>

<em>b)  The margin of error  = 17.3982</em>

<em>c) 95% of confidence intervals are </em>

<em></em>(89.6018 ,124.3982)<em></em>

<em>d) Lower limit of 95% of confidence interval  = 89.6018</em>

<em>upper limit of 95% of confidence interval  = 124.3982</em>

<em>The Population mean is lies between in these intervals</em>

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

Given sample size 'n' = 20

Given sample mean was found to be 107 bpm with a standard deviation of 45 bpm.

<em>Sample mean             </em>x^{-} = 107 bpm<em></em>

<em>Sample standard deviation (S) = 45 bpm</em>

<em>a) standard error of the mean is determined by</em>

<em>     </em>S.E = \frac{S}{\sqrt{n} } = \frac{45}{\sqrt{20} }<em></em>

<em>     S.E = 10.06</em>

<em>b) The margin of error is determined by</em>

<em></em>M.E = \frac{t_{\alpha } X S }{\sqrt{n} }<em></em>

<em>The degrees of freedom  ν   </em>= n-1 =20-1=19<em></em>

<em>   </em>t_{\alpha } = 1.729  

<em></em>M.E = \frac{1.729X 45}{\sqrt{20} }<em></em>

<em></em>M.E = \frac{77.805 }{4.472 } = 17.3982<em></em>

<em>c) 95% of confidence intervals are determined by</em>

<em></em>(x^{-} - t_{\alpha } \frac{S}{\sqrt{n} } , x^{-} + t_{\alpha } \frac{S}{\sqrt{n} } )<em></em>

<em></em>(107 -  1.729\frac{45}{\sqrt{20} } , 107 + 1.729\frac{45}{\sqrt{20} } )<em></em>

<em></em>(107 -  17.3982 } , 107 +17.3982 )<em></em>

<em></em>(89.6018 ,124.3982)<em></em>

<em>d)  </em>

<em>Lower limit of 95% of confidence interval  = 89.6018</em>

<em>upper limit of 95% of confidence interval  = 124.3982</em>

<em>The Population mean is lies between in these intervals</em>

<em></em>

<em></em>

4 0
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