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3 years ago

I have no clue on how to solve this

Mathematics

2 answers:

VMariaS [17]3 years ago

8 0

1, 12 2/3 is your answer

kow [346]3 years ago

8 0

1,12 so your answer

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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

Julli [10]

Answer:

$0.071,1.928$

Step-by-step explanation:

Downtown Store North Mall Store

Sample size n 25 20

Sample mean $\bar{x}$ $9 $8

Sample standard deviation s $2 $1

$n_1=25\\n_2=20$

$\bar{x_1}=9\\ \bar{x_2}=8$

$s_1=2\\s_2=1$

$x_1-x_2=9-8=1$

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

So, z value at 95% confidence interval and 36 degree of freedom = 2.0280

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 $0.071,1.928$

4 0

3 years ago

+ Find the are of the triangle

Yuliya22 [10]

Where’s the picture. Please show it

4 0

3 years ago

PLZ SOMEONE HELP RIGHT AWAY! 10x+2(8x+10)

Umnica [9.8K]

2(13x+10) is the answer

7 0

2 years ago

Read 2 more answers

What is the slope of the line that passed through the points (8,6) and (-2, 10)

lbvjy [14]

Slope: -4/10 = 4/-10

Work: y2-y1/x2-x1

1) 10-6/-2-8

2)4/-10

8 0

3 years ago

Simplify the expression. (3 points) -5+i/2i WILL GIVE BRAINLY

Verizon [17]

$\bf recall~that\qquad i^2=-1\\\\
-------------------------------\\\\
\cfrac{-5+i}{2i}\cdot \cfrac{i}{i}\implies \cfrac{i(-5+i)}{2i^2}\implies \cfrac{-5i+i^2}{2(-1)}\implies \cfrac{-5i+(-1)}{-2}
\\\\\\
\cfrac{-5i-1}{-2}\implies \cfrac{1+5i}{2}$

6 0

3 years ago