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2 years ago

What is 0.8x-x? I will give Brainliest to first answer!!

Mathematics

2 answers:

Yanka [14]2 years ago

6 0

Answer:

-0.2 x

Step-by-step explanation:

=> 0.8x-x

<u>Taking x common</u>

=> x (0.8 - 1)

=> x (-0.2)

=> -0.2 x

geniusboy [140]2 years ago

4 0

Answer:

Explanation:

$0.8x - x$

If a negative term doesn't have a coefficients, it is considered that the coefficient is -1.

$0.8x - 1x$

Calculate:

$= - 0.2x$

Hope this helps..

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What is 218,742 rounded to the nearest hundred thousand?

xxMikexx [17]

Answer:

200,000

Step-by-step explanation:

its easy and simple lol

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3 years ago

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Find the volume of a cylinder made by rotating a rectangle with a width of 2.5 feet and a length of 8 feet as shown in the figur

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<span>the question does not present the picture, but this does not interfere with the resolution</span>

we know that
When a rectangle with a width of 2.5 feet and a length of 8 feet rotates it makes a cylinder with a radius r = 2.5 ft and a height h = 8 ft

volume of a cylinder=pi*r²*h-----> pi*2.5²*8----> 157 ft³

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2 years ago

What value of x is in the solution set of 2(4+2x)>5x+5

Serggg [28]

Answer:

x < 3

Step-by-step explanation:

2(4+2x)>5x+5

Distribute

8 +4x > 5x+5

Subtract 4x from each side

8 +4x-4x > 5x-4x+5

8 > x+5

Subtract 5 from each side

8-5 > x+5-5

3 > x

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2 years ago

Find the value of 2x + 8y when x = - 5 and Y = 6

mestny [16]

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2 years ago

The arrivals of clients at a service firm in Santa Clara is a random variable from Poisson distribution with rate 2 arrivals per

ICE Princess25 [194]

Answer:

1.76% probability that in one hour more than 5 clients arrive

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.

The arrivals of clients at a service firm in Santa Clara is a random variable from Poisson distribution with rate 2 arrivals per hour.

This means that $\mu = 2$

What is the probability that in one hour more than 5 clients arrive

Either 5 or less clients arrive, or more than 5 do. The sum of the probabilities of these events is decimal 1. So

$P(X \leq 5) + P(X > 5) = 1$

We want P(X > 5). So

$P(X > 5) = 1 - P(X \leq 5)$

In which

$P(X \leq 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)$

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

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

$P(X = 1) = \frac{e^{-2}*2^{1}}{(1)!} = 0.2707$

$P(X = 2) = \frac{e^{-2}*2^{2}}{(2)!} = 0.2707$

$P(X = 3) = \frac{e^{-2}*2^{3}}{(3)!} = 0.1804$

$P(X = 4) = \frac{e^{-2}*2^{4}}{(4)!} = 0.0902$

$P(X = 5) = \frac{e^{-2}*2^{5}}{(5)!} = 0.0361$

$P(X \leq 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) = 0.1353 + 0.2702 + 0.2702 + 0.1804 + 0.0902 + 0.0361 = 0.9824$

$P(X > 5) = 1 - P(X \leq 5) = 1 - 0.9824 = 0.0176$

1.76% probability that in one hour more than 5 clients arrive

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2 years ago

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