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Fudgin [204]
3 years ago
5

I need to solve X-9=-12

Mathematics
1 answer:
Rufina [12.5K]3 years ago
7 0

Answer: X = -3

Step-by-step explanation:

X-9=-12

X= -12+9

 =  -3

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Im having a hard time plss help
Dahasolnce [82]
The 2nd one is right
112.5 divided by 15 is 7.5
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3 years ago
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What is an equation of a circle whose center is at (2,-4) and is
Evgesh-ka [11]

The equation of the circle with center at (2,-4) and tangent to the line x = -2 is given by:

(2) (x - 2)² + (y + 4)² = 16

<h3>What is the equation of a circle?</h3>

The equation of a circle of center (x_0, y_0) and radius r is given by:

(x - x_0)^2 + (y - y_0)^2 = r^2

The circle has center at (2,-4), hence x_0 = 2, y_0 = -4.

The center is tangent to the line x = -2, hence it has a point at x = -2, since |-2 - 2| = 4, r = 4 -> r² = 16.

Hence the equation of the circle is given by:

(2) (x - 2)² + (y + 4)² = 16

More can be learned about the equation of a circle at brainly.com/question/24307696

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6 0
2 years ago
How do I factor out the coefficient of the variable such as 1/3b -1/3?
Nutka1998 [239]
I think it is factored out like this:
1/3(b-1)
5 0
3 years ago
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The number of requests for assistance received by a towing service is a Poisson process with rate θ = 4 per hour.a. Compute the
aliya0001 [1]

Answer:

a) 9.93% probability that exactly ten requests are received during a particular 2-hour period

b) 13.53% probability that they do not miss any calls for assistance

c) 2

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 time interval.

Poisson process with rate θ = 4 per hour.

This means that \mu = 4n, in which n is the number of hours.

a. Compute the probability that exactly ten requests are received during a particular 2-hour period.

n = 2, so \mu = 4*2 = 8

This is P(X = 10). So

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

P(X = 10) = \frac{e^{-8}*8^{10}}{(10)!} = 0.0993

9.93% probability that exactly ten requests are received during a particular 2-hour period

b. If the operators of the towing service take a 30-min break for lunch, what is the probability that they do not miss any calls for assistance?

n = 0.5, so \mu = 4*0.5 = 2

This is P(X = 0). So

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

P(X = 0) = \frac{e^{-2}*2^{0}}{(0)!} = 0.1353

13.53% probability that they do not miss any calls for assistance

c. How many calls would you expect during their break?

n = 0.5, so \mu = 4*0.5 = 2

4 0
3 years ago
Please can somebody help me with this question
4vir4ik [10]

Answer:

the answer is=21

Step-by-step explanation:
u have to sum all of its sides
(length+breadth+height).

8 0
2 years ago
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