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vivado [14]
3 years ago
9

Rectangle A′B′C′D′ is the image of rectangle ABCD after a dilation.

Mathematics
1 answer:
Natalija [7]3 years ago
6 0

The scale factor of the dilation from ABCD to A′B′C′D′ is 3.

Explanation:

  • In the smaller rectangle, ABCD the length of the rectangle is 4 - 1 = 3 units, while the width of the rectangle is 5 - 1 = 4 units. For the larger rectangle, A′B′C′D′ the rectangle's length is 12 - 3 = 9 units and the rectangle's width is 15 - 3 = 12 units.
  • To determine the scale factor, we divide the measurement after scaling by the same measurement before scaling. In this case, it can either be the rectangle's length or width.                                                                    So the scale factor = \frac{lengthafterdilation}{lengthbeforedilation}  = \frac{9}{3} = 3.                                                             While the scale factor for width = \frac{widthafterdilation}{widthbeforedilation} = \frac{12}{4} = 3.
  • So the rectangle ABCD is dilated by a scale factor of 3 to produce rectangle A′B′C′D′.

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A change machine changes dollar bills into quarters and nickels. If you receive 12 coins after inserting a dollar bill, how many
Rudik [331]
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3 0
3 years ago
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An apartment complex rents an average of 2.3 new units per week. If the number of apartment rented each week Poisson distributed
masya89 [10]

Answer:

P(X\leq 1) = 0.331

Step-by-step explanation:

Given

Poisson Distribution;

Average rent in a week = 2.3

Required

Determine the probability of renting no more than 1 apartment

A Poisson distribution is given as;

P(X = x) = \frac{y^xe^{-y}}{x!}

Where y represents λ (average)

y = 2.3

<em>Probability of renting no more than 1 apartment = Probability of renting no apartment + Probability of renting 1 apartment</em>

<em />

Using probability notations;

P(X\leq 1) = P(X=0) + P(X =1)

Solving for P(X = 0) [substitute 0 for x and 2.3 for y]

P(X = 0) = \frac{2.3^0 * e^{-2.3}}{0!}

P(X = 0) = \frac{1 * e^{-2.3}}{1}

P(X = 0) = e^{-2.3}

P(X = 0) = 0.10025884372

Solving for P(X = 1) [substitute 1 for x and 2.3 for y]

P(X = 1) = \frac{2.3^1 * e^{-2.3}}{1!}

P(X = 1) = \frac{2.3 * e^{-2.3}}{1}

P(X = 1) =2.3 * e^{-2.3}

P(X = 1) = 2.3 * 0.10025884372

P(X = 1) = 0.23059534055

P(X\leq 1) = P(X=0) + P(X =1)

P(X\leq 1) = 0.10025884372 + 0.23059534055

P(X\leq 1) = 0.33085418427

P(X\leq 1) = 0.331

Hence, the required probability is 0.331

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