If AB = DE CA = FD angle B is congruent to angle E angle A is congruent to angle D , then ∆ABC and ∆DEF are congruent by the ASA
1 answer:
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Answer:
And for this case since the confidence interval contains the value 0 we don't have significant evidence that we have a net change in the levels
Step-by-step explanation:
For this case we have the following info given:
represent the sample size
represent the sample mean
represent the sample deviation
We can calculate the confidence interval for the mean with the following formula:
The confidence level is 0.90 then the significance level would be and the degrees of freedom are given by:
And the critical value for this case would be:
And replacing we got:
And for this case since the confidence interval contains the value 0 we don't have significant evidence that we have a net change or efectiveness in the levels
Answer:
Step-by-step explanation:
So, on substituting all these values, we get
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<h2>ADDITIONAL INFORMATION :-</h2>Sign of Trigonometric ratios in Quadrants
- sin (90°-θ) = cos θ
- cos (90°-θ) = sin θ
- tan (90°-θ) = cot θ
- csc (90°-θ) = sec θ
- sec (90°-θ) = csc θ
- cot (90°-θ) = tan θ
- sin (90°+θ) = cos θ
- cos (90°+θ) = -sin θ
- tan (90°+θ) = -cot θ
- csc (90°+θ) = sec θ
- sec (90°+θ) = -csc θ
- cot (90°+θ) = -tan θ
- sin (180°-θ) = sin θ
- cos (180°-θ) = -cos θ
- tan (180°-θ) = -tan θ
- csc (180°-θ) = csc θ
- sec (180°-θ) = -sec θ
- cot (180°-θ) = -cot θ
- sin (180°+θ) = -sin θ
- cos (180°+θ) = -cos θ
- tan (180°+θ) = tan θ
- csc (180°+θ) = -csc θ
- sec (180°+θ) = -sec θ
- cot (180°+θ) = cot θ
- sin (270°-θ) = -cos θ
- cos (270°-θ) = -sin θ
- tan (270°-θ) = cot θ
- csc (270°-θ) = -sec θ
- sec (270°-θ) = -csc θ
- cot (270°-θ) = tan θ
- sin (270°+θ) = -cos θ
- cos (270°+θ) = sin θ
- tan (270°+θ) = -cot θ
- csc (270°+θ) = -sec θ
- sec (270°+θ) = cos θ
- cot (270°+θ) = -tan θ