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Elena-2011 [213]
4 years ago
10

Lena reflected this figure across the x-axis. She writes the vertices of the image as A'(−2, 8), B'(−5, 6), C'(−8, 8), D'(−4, 2)

. Lena is incorrect because
A) she reflected the pre-image across the y-axis.

B) she reflected the pre-image across the line y = x.

C) she rotated the pre-image 90 degrees counterclockwise.

D) she rotated the pre-image 180 degrees counterclockwise.

Mathematics
1 answer:
garik1379 [7]4 years ago
8 0

Answer:

Option A) she reflected the pre-image across the y-axis

Step-by-step explanation:

we know that

When reflect a point across the the x-axis the rule of the reflection is equal to

(x,y) -----> (x,-y)

When reflect a point across the the y-axis the rule of the reflection is equal to

(x,y) -----> (-x,y)

In this problem we have

A(2,8),B(5,6),C(8,8),D(4,2)

Applying the rule of the reflection across the x-axis

The coordinates of the image would be

A'(2,-8),B'(5,-6),C'(8,-8),D'(4,-2)

Applying the rule of the reflection across the y-axis

The coordinates of the image would be

A'(-2,8),B'(-5,6),C'(-8,8),D'(-4,2)

therefore

she reflected the pre-image across the y-axis.

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Answer:

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Step-by-step explanation:

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The sample space for assigning the two contracts is:

S = {(I, I), (I, II), (I, III), (II, I), (II, II), (II, III), (III, I), (III, II) and (III, III)}

There are total of 9 possible combinations.

So, the probability of selecting any of the combination is, 1/9.

Compute the probability distribution of <em>X</em>₁ and <em>X</em>₂ as follows:

<em>X</em>₁      P (<em>X</em>₁)                      <em>X</em>₂        P (<em>X</em>₂)

0        4/9                        0           4/9

1         4/9                        1            4/9

2         1/9                        2           1/9

Compute the expected values of <em>X</em>₁ and <em>X</em>₂ as follows:

E(X_{1})=\sum X_{1}\cdot P(X_{1})\\\\=(0\times\frac{4}{9})+(1\times\frac{4}{9})+(2\times\frac{1}{9})\\\\=\frac{6}{9}\\\\=\frac{2}{3}               E(X_{2})=\sum X_{2}\cdot P(X_{2})\\\\=(0\times\frac{4}{9})+(1\times\frac{4}{9})+(2\times\frac{1}{9})\\\\=\frac{6}{9}\\\\=\frac{2}{3}

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\text{Total Profit}=\text{Profit}\times [E(X_{1})+E(X_{2})]

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