SAT High School

Graph the image of this figure after a dilation with a scale factor of 3 centered at (−7, −6). Use the polygon tool to graph the dilated figure.

Answers

Answer 1

The dilation transformation gives an image that is similar to the pre-image

but have a different size.

Please find attached the required graph of the image created with MS Excel

Reasons:

Points on the figure are; A(-7, -3), B(-3, -2), C(-4, -5)

The scale factor of dilation = 3

The center of dilation = (-7, -6)

Required:

The graph of the image following the dilation.

Solution:

The given points relative to the center of dilation are;

(-7 - (-7), -3 - (-6)) = (0, 3)

(-3 - (-7), -2 - (-6)) = (4, 4)

(-4 - (-7), -5 - (-6)) = (3, 1)

Following the dilation gives;

3 × (0, 3) = (0, 9)

3 × (4, 4) = (12, 12)

3 × (3, 1) = (9, 3)

Adding the coordinates of the center of dilation to the above relative points of the image gives;

Location of point A' = (0 + (-7) , 9 + (-6)) = (-7, 3)

Location of point B' = (12 + (-7) , 12 + (-6)) = (5, 6)

Location of point C' = (9 + (-7) , 3 + (-6)) = (2, -3)

Therefore, the image is; A'(-7, 3), B'(5, 6), C'(2, -3)

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Graph The Image Of This Figure After A Dilation With A Scale Factor Of 3 Centered At (7, 6). Use The

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Where is the question?

……………………………………

The security system in a house has two units that set off an alarm when motion is detected. Neither one is entirely reliable, but one or both always go off when there is motion anywhere in the house. Suppose that for motion in a certain location, the probability that detector a goes off and detector b does not go off is 0. 25, and the probability that detector a does not go off is 0. 35. What is the probability that detector b goes off?.

Answers

The probability that detector B goes off is 0.75

The probability of an event is known to be the likelihood or chance for an event to occur.

From the given information let consider:

A = the time when detector A goes off, andB = the time when detector B goes off

Since one or both of them always goes off, then:

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∴

Their complements will be zero, i.e.

P(A' and B') = 0

Similarly, we are given that:

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P(B) = P(A) P(B|A) +P(A') P(B|A')

P(B) = P(A) (1 - P(B|A')) + P(A') (1 - P(B'|A')

[tex]\mathbf{P(B) = P(A) (1-\dfrac{P(A \ and \ B')}{P(A)}) +P(A') (1 - \dfrac{P(A' \ and B')}{P(A')})}[/tex]

[tex]\mathbf{P(B) = 0.65 (1-\dfrac{0.25}{0.65}) +0.35 (1 - \dfrac{0}{0.35})}[/tex]

P(B) = 0.75

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EDGE2022

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