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aleksandrvk [35]
3 years ago
6

What ratios are equivalent to 12:3

Mathematics
2 answers:
nlexa [21]3 years ago
8 0

Answer:

4:1

Step-by-step explanation:

adelina 88 [10]3 years ago
5 0

Step-by-step explanation

Ratio equivalent to 12:3

= 12:3

= 12/3 (.°.3x4= 12 and 3x1=3)

= 3x4/3x1 (.°. cancellation of 3 from numerator and denominator)

= 4/1

likewise, 4:1 , 60:15 , 84:21 , 132:33..............etc no. of ratios are equivalent to 12:3

.°. Required ratios which are equivalent to 12:3 are4:1 , 60:15 , 84:21 , 132:33..............etc

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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

P(X⁻ ≥ 96.3) = 0.0087

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} } }

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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)

                      =  1- P( Z<2.383)

                      =  1-( 0.5 -+A(2.38))

                      = 0.5 - A(2.38)

                     = 0.5 -0.4913

                    = 0.0087

<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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3 years ago
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