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Elena L [17]
3 years ago
12

Triangle EFG is dilated by a scale factor of 3 centered at (0, 1) to create triangle E'F'G'. Which statement is true about the d

ilation?

Mathematics
1 answer:
jolli1 [7]3 years ago
5 0

Answer:

Segment EH and segment E prime H prime both pass through the center of dilation (A)

The complete question related to this found on brainly (ID:16812154) is stated below:

Triangle EFG is dilated by a scale factor of 3 centered at (0, 1) to create triangle E'F'G'. Which statement is true about the dilation?

E(0,5) F(1,1) G(-2,1) H(0,1)

a) segment EH and segment E prime H prime both pass through the center of dilation.

b)The slope of segment EF is the same as the slope of segment E prime H prime.

c) segment E prime G prime will overlap segment EG. segment

d) EH ≅ segment E prime H prime.

Step-by-step explanation:

∆EFG = Original image

∆EFG is dilated to give ∆E'F'G'

∆E'F'G' = New image

Scale factor = 3

Center of dilation = (0,1) = H(0,1)

Coordinates ∆EFG : E(0,5) F(1,1) G(-2,1)

To determine the statement that is true about the dilation from the options,

First we would make a diagram on the coordinates of ∆EFG and center of dilation (H).

Find attached the diagram.

Length GF = 3unit

Length G'F' = 3×scale factor = 9unit

Length EH = 4unit

Length E'H' = 4×scale factor = 12unit

Then we would move 3units to the left on same line from G to get the coordinate of G'(mark the point).

Also move 3units to the right on same line from F to get the coordinate of F'(mark the point).

Both of these give length of G'F' = 9unit

Now move 8units to the top from E to get the coordinate of E'(mark the point).

From this you get length E'H' = 12unit

Draw lines connecting the three points to get ∆E'F'G'

See diagram for better understanding

From the diagram, EH and E'H' both pass through (0,1). The other options are wrong.

Therefore, Segment EH and segment E prime H prime both pass through the center of dilation (A)

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A)Mean :10.65

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Mode : 10.7

B)Range = 3.5

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C)The response time of 8.3 minutes should be considered an outlier in comparison to the other response times

Step-by-step explanation:

A)

Data : 11.8 ,10.3, 10.7, 10.6, 11.5, 8.3, 10.5, 10.9, 10.7, 11.2

Mean = \frac{\text{Sum of all observations}}{\text{No. of observations}}\\Mean = \frac{11.8 +10.3+ 10.7+ 10.6+ 11.5+ 8.3+ 10.5+ 10.9+ 10.7+ 11.2}{10}

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Data in ascending order

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10.6

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11.2

11.5

11.8

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Median = \frac{(\frac{n}{2})\text{th term}+(\frac{n}{2}+1)\text{th term}}{2}\\Median = \frac{(\frac{10}{2})\text{th term}+(\frac{10}{2}+1)\text{th term}}{2}\\Median = \frac{10.7+10.7}{2}\\Median = 10.7

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Standard deviation :\sqrt{\frac{(11.8-10.65)^2+(10.3-10.65)^2+......+(10.9-10.65)^2+(10.7-10.65)^2+(11.2-10.65)^2}{10}}

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C)

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For Q3( Median of Upper quartile )

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IQR = Q3-Q1=11.2-10.5=0.7

Range :(Q1-1.5IQR, Q3-1.5IQR)

Range :(10.5-1.5 \times 0.7, 11.2-1.5 \times 0.7)

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