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Aleonysh [2.5K]
3 years ago
15

Solve the equation 2x(2-6)=4(3+1x-2)​

Mathematics
2 answers:
madreJ [45]3 years ago
5 0

Answer:

x =  -  \frac{1}{3}

Step-by-step explanation:

1) Simplify 2 - 6 to -4.

2x  \times  - 4 = 4(3 + 1 \times x - 2)

2) Simplify 1 × x to x.

2x \times  - 4 = 4(3 + x - 2)

3) Simplify 3 + x - 2 to x + 1.

2x \times  - 4 = 4(x + 1)

4) Simplify 2x × -4 to -8x.

- 8x = 4(x + 1)

5) Divide both sides by 4.

- 2x  = x + 1

6) Subtract x from both sides.

- 2x  - x = 1

7) Simplify -2x - x to -3x.

- 3x = 1

8) Divide both sides by -3.

x =  -  \frac{1}{3}

<em><u>Therefor</u></em><em><u>,</u></em><em><u> </u></em><em><u>the</u></em><em><u> </u></em><em><u>answer</u></em><em><u> </u></em><em><u>is</u></em><em><u> </u></em><em><u>x</u></em><em><u> </u></em><em><u>=</u></em><em><u> </u></em><em><u>-1</u></em><em><u>/</u></em><em><u>3</u></em><em><u>.</u></em>

tia_tia [17]3 years ago
3 0

Answer:

no solution

Step-by-step explanation:

if you distribute and simplify:

4x-12 = 12+4x-8

4x-12 = 4x+4

-12 ≠ 4

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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:

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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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