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Aleonysh [2.5K]
2 years ago
7

Which congruence theorem can be used to prove ABR = RCA?

Mathematics
1 answer:
vivado [14]2 years ago
8 0

Answer: The HL Theorem can be used to prove ABR ≅ RCA because both triangles share the same hypotenuse and a leg. The HL theorem states that If the hypotenuse and one leg of a right triangle are congruent to the hypotenuse and one leg of another right triangle, then the triangles are congruent.

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Answer:

x = 9.6

Step-by-step explanation:

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Answer:

z=\frac{0.384-0.362}{\sqrt{0.374(1-0.374)(\frac{1}{4276}+\frac{1}{3908})}}=2.055    

The p value can be calculated from the alternative hypothesis with this probability:

p_v =2*P(Z>2.055)=0.0399    

And the best option for this case would be:

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Step-by-step explanation:

Information provided

X_{1}=1642 represent the number of smokers from the sample in 1995

X_{2}=1415 represent the number of smokers from the sample in 2010

n_{1}=4276 sample from 1995

n_{2}=3908 sample from 2010  

p_{1}=\frac{1642}{4276}=0.384 represent the proportion of smokers from the sample in 1995

p_{2}=\frac{1415}{3908}=0.362 represent the proportion of smokers from the sample in 2010

\hat p represent the pooled estimate of p

z would represent the statistic    

p_v represent the value for the pvalue

System of hypothesis

We want to test the equality of the proportion of smokers and the system of hypothesis are:    

Null hypothesis:p_{1} = p_{2}    

Alternative hypothesis:p_{1} \neq p_{2}    

The statistic is given by:

z=\frac{p_{1}-p_{2}}{\sqrt{\hat p (1-\hat p)(\frac{1}{n_{1}}+\frac{1}{n_{2}})}}   (1)  

Where \hat p=\frac{X_{1}+X_{2}}{n_{1}+n_{2}}=\frac{1642+1415}{4276+3908}=0.374  

Replacing the info given we got:

z=\frac{0.384-0.362}{\sqrt{0.374(1-0.374)(\frac{1}{4276}+\frac{1}{3908})}}=2.055    

The p value can be calculated from the alternative hypothesis with this probability:

p_v =2*P(Z>2.055)=0.0399    

And the best option for this case would be:

C. between 0.01 and 0.05.

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Answer:

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Step-by-step explanation:

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It represents 1/2, 50% or half of the equation.
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