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alex41 [277]
2 years ago
5

HELPPP

Mathematics
1 answer:
ollegr [7]2 years ago
3 0

Answer:

1.25

Step-by-step explanation:

the scale factor is the ratio of corresponding sides, image to original, so

scale factor = \frac{EZ}{TR} = \frac{15}{12} = \frac{5}{4} = 1.25

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In your gym class, your teacher can create teams of 4 for a relay race and have no students left over.The next day, your teacher
Agata [3.3K]

Answer:  20,40,60,80,100

Step-by-step explanation:

The possible  numbers has be the multiples of 5 and 4 in other words the LCM.

The LCM of 5 and 4 are,

20,40,60, 80,100 ....

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3 years ago
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They found 15 seashells ( can you make this a brainliest ansswer pls )
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3 years ago
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What two mixed numbers between 3 and 4 have a product of 9 and 12
dlinn [17]
It's a trick question. There are an infinite number of mixed numbers between 3 and 4 that can multiply to equal 12 (for example, 3 and 3/7 times 3 and 1/2), but there are no mixed numbers between 3 and 4 that can multiply to equal 9. 3 times 3 is not between them but is 3, but that quantity is excluded because 3<x<4. Anything even a small bit above the number 3 would have to be multiplied by 2 and some fraction, which would not be between 3 and 4.
8 0
3 years ago
The probability that a call received by a certain switchboard will be a wrong number is 0.02. Use the Poisson distribution to ap
MAXImum [283]

Answer:

0.2008 = 20.08% probability that among 150 calls received by the switchboard, there are at least two wrong numbers.

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.

The probability that a call received by a certain switchboard will be a wrong number is 0.02.

150 calls. So:

\mu = 150*0.02 = 3

Use the Poisson distribution to approximate the probability that among 150 calls received by the switchboard, there are at least two wrong numbers.

Either there are less than two calls from wrong numbers, or there are at least two calls from wrong numbers. The sum of the probabilities of these events is 1. So

P(X < 2) + P(X \geq 2) = 1

We want to find P(X \geq 2). So

P(X \geq 2) = 1 - P(X < 2)

In which

P(X < 2) = P(X = 0) + P(X = 1)

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) = P(X = 0) + P(X = 1) = 0.0498 + 0.1494 = 0.1992

Then

P(X \geq 2) = 1 - P(X < 2) = 1 - 0.1992 = 0.2008

0.2008 = 20.08% probability that among 150 calls received by the switchboard, there are at least two wrong numbers.

6 0
3 years ago
Helpppp plzzzz It’s due in an hourrrr
adoni [48]

Step-by-step explanation:

1. x+4 =9

x= -4 -9

= -13

2. x-2 =10

x = 2 + 10

x= 12

3. x-9 = -3

x= 9 -3

x=6

4. x +5 = 10

x= -5 -10

x= -15

5. x-6 =13

x = 6 -13

x = -13

6. y-4 = 28

y = 4 +28

y= 32

5 0
3 years ago
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