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sasho [114]
3 years ago
7

-4(2x - 5) – (7 – 3x) = 4(x + 3)

Mathematics
2 answers:
Norma-Jean [14]3 years ago
7 0

Answer:

x  = 1/9.

Explanation:

-4(2x - 5) – (7 – 3x) = 4(x + 3)

-8x + 20 - 7 + 3x = 4x + 12

-8x + 3x - 4x = 12 + 7 - 20

-9x = -1

x = 1/9.

jek_recluse [69]3 years ago
7 0

Answer:

x=1/9

Step-by-step explanation:

-4(2x-5) - (7-3x) = 4(x+3)

x=1/9

-4(2x-5) -(7-3x):  -5x +13

4(x+3):  4x + 12

Subtract 13 from both sides

Simplify: -5x = 4x-1

Subtract 4x from both sides

Simplify -9x = -1

Divide both sides by -9

-9x/-9 = -1/-9

x=1/9

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Answer:

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Step-by-step explanation:

Perimeter of a rectangle = 2(l+b)

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3 0
2 years ago
It is estimated that 0.54 percent of the callers to the Customer Service department of Dell Inc. will receive a busy signal. Wha
stira [4]

Using the binomial distribution, it is found that there is a 0.8295 = 82.95% probability that at least 5 received a busy signal.

<h3>What is the binomial distribution formula?</h3>

The formula is:

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

C_{n,x} = \frac{n!}{x!(n-x)!}

The parameters are:

  • x is the number of successes.
  • n is the number of trials.
  • p is the probability of a success on a single trial.

In this problem:

  • 0.54% of the calls receive a busy signal, hence  p = 0.0054.
  • A sample of 1300 callers is taken, hence n = 1300.

The probability that at least 5 received a busy signal is given by:

P(X \geq 5) = 1 - P(X < 5)

In which:

P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4).

Then:

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 0) = C_{1300,0}.(0.0054)^{0}.(0.9946)^{1300} = 0.0009

P(X = 1) = C_{1300,1}.(0.0054)^{1}.(0.9946)^{1299} = 0.0062

P(X = 2) = C_{1300,2}.(0.0054)^{2}.(0.9946)^{1298} = 0.0218

P(X = 3) = C_{1300,3}.(0.0054)^{3}.(0.9946)^{1297} = 0.0513

P(X = 4) = C_{1300,4}.(0.0054)^{4}.(0.9946)^{1296} = 0.0903

Then:

P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 0.0009 + 0.0062 + 0.0218 + 0.0513 + 0.0903 = 0.1705.

P(X \geq 5) = 1 - P(X < 5) = 1 - 0.1705 = 0.8295

0.8295 = 82.95% probability that at least 5 received a busy signal.

More can be learned about the binomial distribution at brainly.com/question/24863377

#SPJ1

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