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4 years ago

Which sequence of transformations produces an image that is not congruent to the original figure?

2 answers:

6 0

D

Translations, Reflections, and Rotations are transformations that result in a congruent figure.

Dilations on the other hand don't. They change the size of the figure, which means the transformed figure is not congruent to the original.

Vitek1552 [10]4 years ago

5 0

<span>Which sequence of transformations produces an image that is not congruent to the original figure?
A. A reflection across the y-axis followed by a rotation of 180 counterclockwise.
B. A rotation of 90 clockwise followed by a translation of 6 units to the left.
C. A translation of 3 units to the left followed by a reflection across the x-axis.
D. A translation of 5 units to the left followed by a dilation of a factor of 2.</span>

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Answer:

$E(X)=\sum_{i=1}^n X_i P(X_i)$

And if we use the values obtained we got:

$E(X)=5*0.05 +6*0.15 +7*0.33 +8*0.28+ 9*0.12 +10*0.07=7.48$

For this case this value means that the expected score is about 7.48

Step-by-step explanation:

For this case we assume the following probability distribution:

X 5 6 7 8 9 10

P(X) 0.05 0.15 0.33 0.28 0.12 0.07

First we need to find the expected value (first moment) and the second moment in order to find the variance and then the standard deviation.

In order to calculate the expected value we can use the following formula:

$E(X)=\sum_{i=1}^n X_i P(X_i)$

And if we use the values obtained we got:

$E(X)=5*0.05 +6*0.15 +7*0.33 +8*0.28+ 9*0.12 +10*0.07=7.48$

For this case this value means that the expected score is about 7.48

In order to find the standard deviation we need to find first the second moment, given by :

$E(X^2)=\sum_{i=1}^n X^2_i P(X_i)$

And using the formula we got:

$E(X^2)=(5^2 *0.05)+(6^2 *0.15)+(7^2 *0.33)+(8^2 *0.28)+ (9^2 *0.12 +(10^2 *0.07))=57.46$

Then we can find the variance with the following formula:

$Var(X)=E(X^2)-[E(X)]^2 =57.46-(7.48)^2 =1.5096$

And then the standard deviation would be given by:

$Sd(X)=\sqrt{Var(X)}=\sqrt{1.5096}=1.229$

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