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Nastasia [14]
3 years ago
6

12345 what is the answer if you multiply 1x2x3x4x5 and do the same with subtraction, addition and devision, in that order. Then

add the totals together.
Mathematics
1 answer:
Alexandra [31]3 years ago
8 0

Well, we just need to perform the operations:

  • Multiplication: 1\cdot 2 \cdot 3 \cdot 4 \cdot 5 = 120
  • Subtraction: 1-2-3-4-5 = -13
  • Addition: 1+2+3+4+5 = 15
  • Division: 1 \div 2 \div 3 \div 4 \div 5 = \frac{1}{120}

So, if you add all the numbers together you get

120-13+15+\dfrac{1}{120} = 122 + \dfrac{1}{120}

Or, if you prefer,

\dfrac{1681}{120}

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Step By Step Explanation!-

<h2>57.82÷0.784</h2>

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For school lunch students have 3 meat choices (beef, chicken, or fish), 2 potato choices (mashed or fried), and 3 drink choices
ioda

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18

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A quality control expert at Glotech computers wants to test their new monitors. The production manager claims they have a mean l
tankabanditka [31]

Answer:

The probability that the mean monitor life would be greater than 96.3 months in a sample of 84 monitors

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Step-by-step explanation:

<u><em>Step(i):-</em></u>

Given that the mean of the Population = 95

Given that the standard deviation of the Population = 5

Let 'X' be the random variable in a normal distribution

Let X⁻ = 96.3

Given that the size 'n' = 84 monitors

<u><em>Step(ii):-</em></u>

<u><em>The Empirical rule</em></u>

         Z = \frac{x^{-} -mean}{\frac{S.D}{\sqrt{n} } }

        Z = \frac{96.3 - 95}{\frac{5}{\sqrt{84} } }

       Z = 2.383

The probability that the mean monitor life would be greater than 96.3 months in a sample of 84 monitors

    P(X⁻ ≥ 96.3) = P(Z≥2.383)

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<u><em>Final answer:-</em></u>

The probability that the mean monitor life would be greater than 96.3 months in a sample of 84 monitors

P(X⁻ ≥ 96.3) = 0.0087

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Step-by-step explanation:

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