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nirvana33 [79]
4 years ago
14

Plot C(-1,–7), D (6,–7), E(5,3), and F(5,-5) in a coordinate plane. Are CD and EF congruent?

Mathematics
1 answer:
N76 [4]4 years ago
7 0

Answer:

No, they are not congruent

Step-by-step explanation:

By CD and EF being congruent, we want to know if they are equal.

What we can simply do here is to find the distance between the points present.

That is, the distance between C and D, the distance between E and F

To get the distances between two points , we use the following equation

D = √(x2-x1)^2 + (y2-y1)^2

For CD, we have

D = √(6 -(-1)^2 + (-7 + 7)^2

D = √(7)^2 = 7 units

For EF, we have

D = √(5-5)^2 + (-5-3)^2

D = √(-8)^2

D = √(64) = 8 units

This clearly shows that they are both not congruent

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Answer:

The 95% confidence interval would be given by -5.980 \leq \mu_1 -\mu_2 \leq -0.720  

p_v =2*P(Z

Step-by-step explanation:

Previous concepts  

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

\bar X_1 =65.2 represent the sample mean 1

\bar X_2 =68.55 represent the sample mean 2

n1=10 represent the sample 1 size  

n2=10 represent the sample 2 size  

s_1 =3.55 sample standard deviation for sample 1

s_2 =2.29 sample standard deviation for sample 2

\sigma =3 represent the population standard deviation

\mu_1 -\mu_2 parameter of interest.

Part a

The confidence interval for the difference of means is given by the following formula:  

(\bar X_1 -\bar X_2) \pm z_{\alpha/2}\sqrt{\sigma^2(\frac{1}{n_1}+\frac{1}{n_2})} (1)  

The point of estimate for \mu_1 -\mu_2 is just given by:

\bar X_1 -\bar X_2 =65.2-68.55=-3.35

Since the Confidence is 0.95 or 95%, the value of \alpha=0.05 and \alpha/2 =0.025, and we can use excel, a calculator or a tabel to find the critical value. The excel command would be: "=-NORM.INV(0.025,0,1)".And we see that z_{\alpha/2}=1.96  

The standard error is given by the following formula:

SE=\sqrt{\sigma^2(\frac{1}{n_1}+\frac{1}{n_2})}

And replacing we have:

SE=\sqrt{3^2(\frac{1}{10}+\frac{1}{10})}=1.342

Confidence interval

Now we have everything in order to replace into formula (1):  

-3.35-1.96\sqrt{3^2(\frac{1}{10}+\frac{1}{10})}=-5.980  

-3.35+1.96\sqrt{3^2(\frac{1}{10}+\frac{1}{10})}=-0.720  

So on this case the 95% confidence interval would be given by -5.980 \leq \mu_1 -\mu_2 \leq -0.720  

If the system of hypothesis are given by:

Null Hypothesis:\mu_1 -\mu_2=0

Alternative hypothesis:\mu_1 -\mu_2 \neq 0

The statistic would be:

z=\frac{\bar X_1 -\bar X_2 -0}{\sqrt{\sigma^2(\frac{1}{n_1}+\frac{1}{n_2})}}

And if we replace we got:

z=\frac{65.2 -68.55 }{\sqrt{3^2(\frac{1}{10}+\frac{1}{10})}}=-2.497

And the p value would be given by:

p_v =2*P(Z

And with 5% of significance we have enough evidence to reject the null hypothesis since the p_v < \alpha

5 0
3 years ago
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