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2 years ago

Describe the relationship between the point D(6, 9) and the point A(8, 12) in terms of dilations.

Mathematics

1 answer:

Irina18 [472]2 years ago

6 0

Answer:

one is smaller then the other i think thats it

Step-by-step explanation:

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the morgan famili and the alexander family each ised their sprinkles last summer. the water output was 20. l per hour. The water

erik [133]

Answer:

number of hours Alexander family ‘s sprinkler work = 20

number of hours Morgan family ‘s sprinkler work = 30

Step-by-step explanation:

Given in the question,

water output rate for the Alexander family ‘s sprinkler = 40 L

water output rate for the Morgan family ‘s sprinkler = 20 L

Suppose, number of hours Alexander family ‘s sprinkler work = x

Suppose, number of hours Morgan family ‘s sprinkler work = y

Equation1

x + y = 50

Equation 2

40x + 20y = 1400

x = 50 - y

Put this value of x in equation 2

40(50-y) + 20y = 1400

2000 - 40y + 20y = 1400

-20y = 1400 - 2000

-20y = -600

y = 30

Put this value of y in equation 1 to get the value of x

x + 30 = 50

x = 50 - 30

x = 20

5 0

3 years ago

What is Something divided by something equals 140

inn [45]

Answers my vary but what i got is 140 divided by 10 = 14

4 0

3 years ago

Read 2 more answers

2.8 is 4% of what number

AnnZ [28]

70 you solve by making a proportion 2.8/x =4/100

4 0

4 years ago

The number of typing errors made by a typist has a Poisson distribution with an average of three errors per page. If more than t

damaskus [11]

Answer:

0.6472 = 64.72% probability that a randomly selected page does not need to be retyped.

Step-by-step explanation:

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 interval.

Poisson distribution with an average of three errors per page

This means that $\mu = 3$

What is the probability that a randomly selected page does not need to be retyped?

Probability of at most 3 errors, so:

$P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)$

In which

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

$P(X = 0) = \frac{e^{-3}*3^{0}}{(0)!} = 0.0498$

$P(X = 1) = \frac{e^{-3}*3^{1}}{(1)!} = 0.1494$

$P(X = 2) = \frac{e^{-3}*3^{2}}{(2)!} = 0.2240$

$P(X = 3) = \frac{e^{-3}*3^{3}}{(3)!} = 0.2240$

Then

$P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.0498 + 0.1494 + 0.2240 + 0.2240 = 0.6472$

0.6472 = 64.72% probability that a randomly selected page does not need to be retyped.

3 0

3 years ago

Which of these equations have no solution? Check all that apply.

ollegr [7]

Answer:

the first one

Step-by-step explanation:

6 0

3 years ago

Read 3 more answers