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liq [111]
3 years ago
15

Find x and y on triangleAlso the degree is 30 and the other thing is 7sqrt3​

Mathematics
1 answer:
nydimaria [60]3 years ago
6 0

Answer:

y =7

x =14

Step-by-step explanation:

Since this is a right triangle we can use trig functions

tan 30 = opp /adj

tan 30 = y/ 7 sqrt(3)

7 sqrt(3)  tan 30 = y

7 sqrt(3) * 1/ sqrt(3) =t

7 =y

sin 30 = opp/ hyp

sin 30 = 7/x

x sin 30 =7

x = 7/ sin 30

x = 7 / 1/2

x = 14

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In order to estimate the difference between the average hourly wages of employees of two branches of a department store, the fol
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Sample mean \bar{x}                         $9                        $8

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Standard error of difference of means = \sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}

Standard error of difference of means = \sqrt{\frac{2^2}{25}+\frac{1^2}{20}}

Standard error of difference of means = 0.458

Degree of freedom = \frac{\sqrt{(\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}})^2}{\frac{(\frac{s_1^2}{n_1})^2}{n_1-1}+\frac{(\frac{s_2^2}{n_2})^2}{n_2-1}}

Degree of freedom = \frac{\sqrt{(\frac{2^2}{25}+\frac{1^2}{20}})^2}{\frac{(\frac{2^2}{25})^2}{25-1}+\frac{(\frac{1^2}{20})^2}{20-1}}

Degree of freedom =36

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Confidence interval = (x_1-x_2)-z \times SE(x_1-x_2),(x_1-x_2)+z \times SE(x_1-x_2)

Confidence interval = 1-(2.0280)\times 0.458,1+(2.0280)\times 0.458

Confidence interval = 0.071,1.928

Hence Option A is true

Confidence interval is  0.071,1.928

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