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Scilla [17]
2 years ago
7

Suppose the figure is dilated by 3/2 with the center of dilation at the origin. Which would be the ordered pairs for the image o

f ABCD?
Group of answer choices

A’(-3, 4.5), B’(-6,-6), C’(6, 0), D’(7.5, 6)

A’(-4/3, 2), B’(10/3, 8/3), C’(8/3, 0), D’(-8/3, -8/3)

A’(-4, 6), B’(-8, -8), C’(8, 0), D’(10,8)

A’(-6, 9), B’(-12, -12), C’(12,0), D’(15, 12)

Mathematics
1 answer:
Mandarinka [93]2 years ago
3 0

Answer:

D

Step-by-step explanation: just took the quiz

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is this good?



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3 years ago
Use the drop-down menu to complete the statement.
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Answer:

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Step-by-step explanation:

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3 years ago
Let X equal the number of typos on a printed page with a mean of 4 typos per page.
timama [110]

Answer:

a) There is a 98.17% probability that a randomly selected page has at least one typo on it.

b) There is a 9.16% probability that a randomly selected page has at most one typo on it.

Step-by-step explanation:

Since we only have the mean, we can solve this problem by a Poisson distribution.

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}

In which

x is the number of sucesses

e = 2.71828 is the Euler number

\mu is the mean in the given time interval.

In this problem, we have that \mu = 4

(a) What is the probability that a randomly selected page has at least one typo on it?

Thats is P(X \geq 1). Either a number is greater or equal than 1, or it is lesser. The sum of the probabilities must be decimal 1. So:

P(X < 1) + P(X \geq 1) = 1

P(X \geq 1) = 1 - P(X < 1)

In which

P(X < 1) = P(X = 0).

So

P(X = 0) = \frac{e^{-4}*4^{0}}{(0)!} = 0.0183

P(X \geq 1) = 1 - P(X < 1) = 1 - 0.0183 = 0.9817

There is a 98.17% probability that a randomly selected page has at least one typo on it.

(b) What is the probability that a randomly selected page has at most one typo on it?

This is P = P(X = 0) + P(X = 1). So:

P(X = 0) = \frac{e^{-4}*4^{0}}{(0)!} = 0.0183

P(X = 1) = \frac{e^{-4}*4^{1}}{(1)!} = 0.0733

P = P(X = 0) + P(X = 1) = 0.0183 + 0.0733 = 0.0916

There is a 9.16% probability that a randomly selected page has at most one typo on it.

3 0
3 years ago
Suppose a teacher finds that the
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Answer:

If a student with 2 absences got a score of 12, we can affirm that the student got 6 points for every absence she or he had.

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Step-by-step explanation:

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3 years ago
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