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olchik [2.2K]
3 years ago
10

Quadrilateral W X Y Z is shown. Diagonals are drawn from point W to point Y and from point Z to point X and intersect at point C

. The lengths of W C and C Y are congruent. Which best explains if quadrilateral WXYZ can be a parallelogram? WXYZ is a parallelogram because diagonal XZ is bisected. WXYZ is not necessarily a parallelogram because it is unknown if CZ = CY. WXYZ is a parallelogram because ZC + CX = ZX. WXYZ is not necessarily a parallelogram because it is unknown if WC = CY.
Mathematics
2 answers:
podryga [215]3 years ago
7 0

Answer:

3 the answer is three on edgenuity

Step-by-step explanation

u    u

 w

kkurt [141]3 years ago
5 0

Answer:

A.  WXYZ is a parallelogram because diagonal XZ is bisected.

Step-by-step explanation:

A parallelogram is an example of quadrilaterals which has pairs of opposite sides parallel and equal.

In the given question;

WC ≅ CY

⇒ WY = WC + CY

Considering diagonal WY,

ΔWYZ ≅ ΔWYX (diagonal property of a parallelogram)

WY ⊥ XZ (diagonals of a parallelogram bisect each other)

Thus;

XC ≅ CZ

Therefore. WXYZ is a parallelogram.  

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katrin [286]

Answer:

x= 105

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Step-by-step explanation:

3y=4y-40 because they are alternate interior

Solve for y.

y=40

4y-40=x+15 because they are vertical angles.

Substitute 40 in for y.

4(40)-40=x+15

Solve.

x=105

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I need the answer for these.<br> 1a: <br> 1b:<br> 1c:
Fantom [35]

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What fraction makes the equation true?<br> Enter the answer in the boxes.<br><br> 5/6x =20/24
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Answer:

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Step-by-step explanation:

To find the value of x, we have to move all the real numbers to the other side.

5/6x = 20/24

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Suppose each of the following data sets is a simple random sample from some population. For each dataset, make a normal QQ plot.
adell [148]

Answer:

a) For this case the histogram is not too skewed and we can say that is approximately symmetrical so then we can conclude that this dataset is similar to a normal distribution

b) For this case the data is skewed to the left and we can't assume that we have the normality assumption.

c) This last case the histogram is not symmetrical and the data seems to be skewed.

Step-by-step explanation:

For this case we have the following data:

(a)data = c(7,13.2,8.1,8.2,6,9.5,9.4,8.7,9.8,10.9,8.4,7.4,8.4,10,9.7,8.6,12.4,10.7,11,9.4)

We can use the following R code to get the histogram

> x1<-c(7,13.2,8.1,8.2,6,9.5,9.4,8.7,9.8,10.9,8.4,7.4,8.4,10,9.7,8.6,12.4,10.7,11,9.4)

> hist(x1,main="Histogram a)")

The result is on the first figure attached.

For this case the histogram is not too skewed and we can say that is approximately symmetrical so then we can conclude that this dataset is similar to a normal distribution

(b)data = c(2.5,1.8,2.6,-1.9,1.6,2.6,1.4,0.9,1.2,2.3,-1.5,1.5,2.5,2.9,-0.1)

> x2<- c(2.5,1.8,2.6,-1.9,1.6,2.6,1.4,0.9,1.2,2.3,-1.5,1.5,2.5,2.9,-0.1)

> hist(x2,main="Histogram b)")

The result is on the first figure attached.

For this case the data is skewed to the left and we can't assume that we have the normality assumption.

(c)data = c(3.3,1.7,3.3,3.3,2.4,0.5,1.1,1.7,12,14.4,12.8,11.2,10.9,11.7,11.7,11.6)

> x3<-c(3.3,1.7,3.3,3.3,2.4,0.5,1.1,1.7,12,14.4,12.8,11.2,10.9,11.7,11.7,11.6)

> hist(x3,main="Histogram c)")

The result is on the first figure attached.

This last case the histogram is not symmetrical and the data seems to be skewed.

7 0
3 years ago
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