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True [87]
3 years ago
14

Zahid is paid a set wage of 774.72 for a 36 hours week. plus time and a half for overtime. In one particular week, he worked 43

hours. what were his earning that week?
Mathematics
1 answer:
Viktor [21]3 years ago
5 0

Answer:

1162.08

Step-by-step explanation:

zahid is paid 774.72 for 36 hrs so,

he worked 43 hrs,

so no. of hours ot: 43-36

7 hrs

now we will find the half of 774.72

387.36

now add 774.72+387.36

1162.08

so that was his earning

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What is the difference of the fractions? Use the number line to help find the answer.
vova2212 [387]
<h2>Explanation:</h2>

Hello, remember you need to write complete questions in order to get good and exact answers. Here you haven't provided any fractions, so I'll give you my own fractions.

The first fraction is:

\frac{1}{2}

The second fraction is:

\frac{1}{4}

So let's say that difference is:

\frac{1}{2}-\frac{1}{4}

Therefore, the result is:

\frac{1}{2}-\frac{1}{4}=\frac{1}{4}

The representation of this problem is shown using the number line below. As you can see, we have written both 1/2 and 1/4 and the difference is also indicated giving the result 1/4. That is, if we walk from 1/4 to 1/2 we'll walk 1/4 units.

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4 years ago
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Please help Which expression is equal to...?
Margarita [4]

Answer:

the last one -\sqrt{5}

Step-by-step explanation:

3 0
3 years ago
Find the percent increase or decrease: 43 to 78
Kobotan [32]

Answer:

81.39% increase.

Step-by-step explanation:

Given: There is change in number from 43 to 78.

First lets find the amount of change or difference in number.

Difference in number= 78-43= 35

∴ Difference in number show that there is an increase of 35 number.

Now, finding the percent change in the number.

Percent= \frac{Difference\ in\ number}{base\ number} \times 100

⇒ Percent change in number= \frac{35}{43} \times 100= 81.39\%

∴ 81.39% increase in the number.

3 0
3 years ago
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

3 0
3 years ago
Use systems of equations to solve the problem.
S_A_V [24]
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