∆ABC is rotated 100° counterclockwise about point P to create ∆A'B'C'. what is angle m C'A'B'?
2 answers:
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Answer:
If we compare the p value and the significance level assumed we see that
so we can conclude that we have enough evidence to fail reject the null hypothesis, so we can't conclude that the height of men actually its NOT significant higher than 0.3 at 1% of signficance.
Step-by-step explanation:
Data given and notation
represent the sample mean
represent the sample standard deviation for the sample
sample size
represent the value that we want to test
represent the significance level for the hypothesis test.
t would represent the statistic (variable of interest)
represent the p value for the test (variable of interest)
State the null and alternative hypotheses.
We need to conduct a hypothesis in order to check if the mean is higher than 3, the system of hypothesis would be:
Null hypothesis:
Alternative hypothesis:
If we analyze the size for the sample is > 30 and we know the population deviation so is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:
(1)
z-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".
Calculate the statistic
We can replace in formula (1) the info given like this:
P-value
Since is a one side test the p value would be:
Conclusion
If we compare the p value and the significance level assumed we see that
so we can conclude that we have enough evidence to fail reject the null hypothesis, so we can't conclude that the height of men actually its NOT significant higher than 0.3 at 1% of signficance.
CE ⊥ CF because m∠ECA + m∠FCA = 90°
Step-by-step explanation:
In ∆ABC
- On the extension of the side BC , draw a line segment CD ≅ CA
- Draw the segment AD
- The line segment CE is the angle bisector of ∠ACB
- The line segment CF is the median towards AD in ∆ ACD
We want to prove that CF ⊥ CE
Look to the attached figure
In Δ ABC
∵ CE is the bisector of angle ACB
∴ ∠ACE ≅ ∠BCE
In Δ ACD
∵ CA = CD
∴ Δ ACD is an isosceles triangle
∵ AD is the median towards AD
- In any isosceles triangle the median from a vertex to its opposite
side bisects this vertex
∴ AD bisects ∠ACD
∴ ∠ACF ≅ ∠DCF
∵ BCD is a straight segment
∵ CE , CA , CF are drawn from point C
∴ m∠BCE + m∠ACE + m∠ACF + m∠DCF = 180°
∵ m∠ACE ≅ m∠BCE
∵ m∠ACF ≅ m∠DCF
- Replace m∠BCE by m∠ACE and m∠DCF by m∠ACF
∴ m∠ACE + m∠ACE + m∠ACF + m∠ACF = 180°
∴ 2 m∠ACE + 2 m∠ACF = 180°
- Divide all terms by 2
∴ m∠ACE + m∠ACF = 90°
∴ EC ⊥ CF
CE ⊥ CF because m∠ECA + m∠FCA = 90°
Learn more:
You can learn more about perpendicular lines in brainly.com/question/11223427
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