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2 years ago

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Quadrilateral PERK has vertices at P(4,4), E(12,4), R(10,−2), and K(2,−2). Find the vertices of the quadrilateral after a dilati

on with scale factor 32.
These are the options. 1,2,3, and 4.

1. P'(8/3, 8/3), E'(8, 8/3), R'(20/3, −4/3), and K'(4/3, −4/3)
2. P'(−6, −6), E'(−18, −6), R'(−15, 3), and K'(−3, 3)
3. P'(6, 6), E'(18, 6), R'(15, −3), and K'(3, −3)
4. P'(−8/3, −8/3), E'(−8, −8/3), R'(−20/3, 4/3), and K'(−4/3, 4/3)

1 answer:

Murrr4er [49]2 years ago

8 0

Answer:

option 3

Step-by-step explanation:

Dilation by a scale factor of 3/2 multiplies each of the coordinate values by 3/2. For example, P' = (3/2)P = (3/2)(4, 4) = (6, 6).

The only choice with coordinates matching these is option 3.

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$(d)\ 0.936$

Step-by-step explanation:

Given

$p \to$ close

$q \to$ fail to close

$p = 60\%$

$p = 0.60$

First, calculate the value of q

Using complement rule

$q = 1 - p$

$q = 1 - 0.60$

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So, we have:

$p = 0.60$ and $q = 0.40$

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This means that he closes the first three attempts. The event is represented as: p p p q

So, we have:

$Pr = p*p*p*q$

$Pr = p^3*q$

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$Pr = 0.0864$

Solving (b): He closes for the first time on the 3rd attempt

This means that he fails to close the first two attempts. The event is represented as: q q p

So, we have:

$Pr = q * q * p$

$Pr = q^2 * p$

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$Pr = 0.096$

Solving (c): First he closes is his 2nd attempt

This means that he fails to close the first. The event is represented as: q p

So, we have:

$Pr = q * p$

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$Pr = 0.24$

Solving (d): The first he close is one of his 3 attempts

To do this, we make use of complement rule

The event that he does not close any of his first three attempts is: q q q

The probability is:

$Pr = q*q*q$

$Pr = q^3$

The opposite is that the first he closes is one of the first three

So, we have:

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$Pr' = 1- q^3$

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