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Agata [3.3K]
2 years ago
10

Given the three vertices W(−1, 4), X(−3, −2), and Y(3, −4), what are the coordinates of Z that make quadrilateral WXYZ a square?

(4 points)
Mathematics
1 answer:
prohojiy [21]2 years ago
5 0

Answer:

I don't see how the three existing points could ever become a square with the addition of a foiurth point.

Step-by-step explanation:

See the attached image.

A square would require that all angles be 90 degrees.  The given points are the top three points on the graph.  If we enter the two equations that intersect these points (blue and black lines), we can see that the angle on top is not 90 degrees.  I can't see that this could ever be a square with a fourth point, z.  I did find a value for z that make the four points a parallelogram.

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Refer to the accompanying data set and construct a ​% confidence interval estimate of the mean pulse rate of adult​ females; the
ivolga24 [154]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The 95% confidence interval of mean pulse rate for adult female is 71.98 <  \mu <  79.88

The 95% confidence interval of mean pulse rate for adult male is  62.89 <  \mu <  70.57

The correct option is  C

Step-by-step explanation:

  Generally the sample mean for Male pulse rate is mathematically represented as

       \= x_1   = \frac{\sum x_i }{n}

= >    \= x_1   = \frac{81 + 74 + \cdots + 59 }{40 }

= >    \= x_1   = 66.73

Generally the standard deviation for male pulse rate is mathematically represented as

      \sigma_1 = \sqrt{\frac{\sum (x - \= x)^2 }{n-1 } }

=>   \sigma_1 = \sqrt{\frac{\sum (81 - 66.73 )^2 + (74 - 66.73 )^2+ \cdot + (59 - 66.73 )^2 }{40-1 } }

=>   \sigma_1 = 12.24

From the question we are told the confidence level is  95% , hence the level of significance is    

      \alpha = (100 - 95 ) \%

=>   \alpha = 0.05

Generally from the normal distribution table the critical value  of  \frac{\alpha }{2} is  

   Z_{\frac{\alpha }{2} } =  1.96

Generally the margin of error is mathematically represented as  

      E = Z_{\frac{\alpha }{2} } *  \frac{\sigma_1 }{\sqrt{n} }

=>    E = 1.96 *  \frac{12.4 }{\sqrt{40} }

=>    E =3.84

Generally 95% confidence interval is mathematically represented as  

      \= x_1 -E <  \mu <  \=x_1  +E

=>    66.73 -3.84 <  \mu < 66.73 +3.84

=>    62.89 <  \mu <  70.57

  Generally the sample mean for Female pulse rate is mathematically represented as

       \= x_2   = \frac{\sum x_i }{n}

= >    \= x_2   = \frac{81 + 94 + \cdots + 73 }{40 }

= >    \= x_2   = 75.93

Generally the standard deviation for Female  pulse rate is mathematically represented as

      \sigma_2 = \sqrt{\frac{\sum (x - \= x)^2 }{n-1 } }

=>   \sigma_2 = \sqrt{\frac{\sum (81 - 66.73 )^2 + (94 - 66.73 )^2+ \cdot + (73 - 66.73 )^2 }{40 } }

=>   \sigma_2 = 12.73

Generally from the normal distribution table the critical value  of  \frac{\alpha }{2} is  

   Z_{\frac{\alpha }{2} } =  1.96

Generally the margin of error is mathematically represented as  

      E = Z_{\frac{\alpha }{2} } *  \frac{\sigma_2 }{\sqrt{n} }

=>    E = 1.96 *  \frac{12.73 }{\sqrt{40} }

=>    E =3.95

Generally 95% confidence interval is mathematically represented as  

      \= x_2 -E <  \mu <  \=x_2  +E

=>    75.93 -3.95 <  \mu < 75.93 + 3.95

=>    71.98 <  \mu <  79.88

8 0
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John took a math test and got 19 out of 25 questions correct. What percent of the questions did he answer correctly ?. Single ch
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Answer:

76%

Step-by-step explanation:

  • Start with the fraction \frac{19}{25}
  • Get the denominator to 100 by multiplying both the numerator ad denominator by 4
  • This will give you the fraction \frac{76}{100}
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What is the volume of the figure???
lianna [129]

Answer:

1530cm

Step-by-step explanation:

First figure      (9x10)x9=810

Second figure (18x10)x4=720

720+810=1530

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Put the following in order for the most area in the tails of the distribution. ​(a) Standard Normal Distribution ​(b) Student's​
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Answer:

(b) (c) (a)

Step-by-step explanation:

Standard Normal distribution has a higher peak in the center, with more area in this región, hence it has less area in its tails.

Student's​ t-Distribution has a shape similar to the Standard Normal Distribution, with the difference that the shape depends on the degree of freedom. When the degree of freedom is smaller the distribution becomes flatter, so it has more area in its tails.

Student's​ t-Distributionwith 1515 degrees of freedom has mores area in the tails than the Student's​ t-Distribution with 2020 degrees of freedom and the latter has more area than Standard Normal Distribution

5 0
3 years ago
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