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LUCKY_DIMON [66]
2 years ago
15

Which of the following are reasons used in the proof that the angle-bisector

Mathematics
1 answer:
Darya [45]2 years ago
7 0

The angle-bisector construction can be used to bisect any angle and this is proven by:

  • A. All of the radii of a circle are congruent.
  • B. SSS triangle congruence postulate.
  • D. CPCTC.
<h3 /><h3>What is the proof that the angle-bisector construction can bisect any angle?</h3>

The angle-bisector construction means that a line is made that can divide an angle into two congruent angles.

This is shown by the fact that all the radii of a circle are congruent and that the corresponding parts of congruent triangles are also congruent (CPCTC) when an angle bisector is made on a circle or triangle respectively.

Find out more on angle bisectors at brainly.com/question/24334771.

#SPJ1

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Alex_Xolod [135]

Answer:

(6,4)

Step-by-step explanation:

Midpoint=((x1+x2)/2 ,(y1+y2)/2)

((10+2)/2 ,(7+1)/2)

(6,4)

x1=10

x2=2

y1=2

y2=1

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The SOLT missionaries are throwing a birthday party. There is obviously cake. Mr. Daniel only wants 1/2 of a piece. Mr. Richie e
marishachu [46]

Answer:

1/ 112

Step-by-step explanation:

1-½ = ½ ( Mr. Daniel's leftover)

½×½= ¼ ( Mr. Richie)

½×¼= 1/8 ( Ms Laura)

½×1/8 = 1/16 ( Ms. Kimberly)

1/7 × 1/16 = 1/ 112 (the last turn)

so, Ms. Sarah eats 1/ 112 of a piece.

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3 years ago
We are conducting a hypothesis test to determine if fewer than 80% of ST 311 Hybrid students complete all modules before class.
Juli2301 [7.4K]

Answer:

z=\frac{0.881 -0.8}{\sqrt{\frac{0.8(1-0.8)}{110}}}=2.124  

Null hypothesis:p\geq 0.8  

Alternative hypothesis:p < 0.8  

Since is a left tailed test the p value would be:  

p_v =P(Z  

So the p value obtained was a very high value and using the significance level assumed \alpha=0.05 we have p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis.

Be Careful with the system of hypothesis!

If we conduct the test with the following hypothesis:

Null hypothesis:p\leq 0.8  

Alternative hypothesis:p > 0.8

p_v =P(Z>2.124)=0.013  

So the p value obtained was a very low value and using the significance level assumed \alpha=0.05 we have p_v so we can conclude that we have enough evidence to reject the null hypothesis.

Step-by-step explanation:

1) Data given and notation  

n=110 represent the random sample taken

X=97 represent the students who completed the modules before class

\hat p=\frac{97}{110}=0.882 estimated proportion of students who completed the modules before class

p_o=0.8 is the value that we want to test

\alpha represent the significance level

z would represent the statistic (variable of interest)

p_v{/tex} represent the p value (variable of interest)  2) Concepts and formulas to use  We need to conduct a hypothesis in order to test the claim that the true proportion is less than 0.8 or 80%:  Null hypothesis:[tex]p\geq 0.8  

Alternative hypothesis:p < 0.8  

When we conduct a proportion test we need to use the z statistic, and the is given by:  

z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}} (1)  

The One-Sample Proportion Test is used to assess whether a population proportion \hat p is significantly different from a hypothesized value p_o.

3) Calculate the statistic  

Since we have all the info requires we can replace in formula (1) like this:  

z=\frac{0.881 -0.8}{\sqrt{\frac{0.8(1-0.8)}{110}}}=2.124  

4) Statistical decision  

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.  

The significance level assumed \alpha=0.05. The next step would be calculate the p value for this test.  

Since is a left tailed test the p value would be:  

p_v =P(Z  

So the p value obtained was a very high value and using the significance level assumed \alpha=0.05 we have p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis.

Be Careful with the system of hypothesis!

If we conduct the test with the following hypothesis:

Null hypothesis:p\leq 0.8  

Alternative hypothesis:p > 0.8

p_v =P(Z>2.124)=0.013  

So the p value obtained was a very low value and using the significance level assumed \alpha=0.05 we have p_v so we can conclude that we have enough evidence to reject the null hypothesis.

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Evaluate 9P2- need to make a good grade
Svetllana [295]
81, if you mean 9 Squared.
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What is an expression equivalent to 7(c-4)
Lemur [1.5K]

Answer:

7c - 28

Step-by-step explanation:

7(c - 4)

(7 × c) + (7 × - 4)

7c + (7 × - 4)

7c + -28

7c - 28

7 0
3 years ago
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