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Katena32 [7]
3 years ago
5

Given the function, d(t) = 50t, the variable t represents which of the following? Select all that apply. input output function i

ndependent variable dependend variable
will give brainlyist if you want it
Mathematics
2 answers:
34kurt3 years ago
6 0

Answer:

Input

Independent variable

Step-by-step explanation:

we know that

<u><em>Independent variables</em></u>, are the values that can be changed or controlled in a given model or equation

<u><em>Dependent variables</em></u>, are the values that result from the independent variables

we have the function

d(t)=50t

In this problem

The function d(t) represent the dependent variable or the output

The variable t represent the independent variable or input

taurus [48]3 years ago
5 0

Answer:

hmmm

Step-by-step explanation:

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Answer:

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Step-by-step explanation:

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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} } }

       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)

                      =  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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