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Drupady [299]
3 years ago
7

Six students are working on math. Two-thirds of them are working on fractions. How many students are working on fractions?

Mathematics
1 answer:
dybincka [34]3 years ago
3 0
(2/3)- students working on math
6-total students
(2/3) of 6 students+= (2/3)x6 = (12/3) = 4
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horsena [70]

Answer:

The mean oil-change time that would there be a 10% chance of being at or below is 15.46 minutes.

Step-by-step explanation:

We are given that records indicate that the mean time is 16.2 minutes, and the standard deviation is 3.4 minutes.

On a typical​ Saturday, the​ oil-change facility will perform 35 oil changes between 10 A.M. and 12 P.M. Assuming that data follows normal distribution.

Let \bar X = <u><em>sample mean time</em></u>

The z score probability distribution for sample mean is given by;

                          Z  =  \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }  ~ N(0,1)

where, \mu = population mean time = 16.2 minutes

            \sigma = standard deviation = 3.4 minutes

            n = sample size = 35 oil changes

<u>Now, the mean oil-change time that would there be a 10% chance of being at or below is given by;</u>

<u></u>

          P(X \leq x) = 0.10          {where x is required mean oil-change time}

          P( \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } } \leq \frac{x-16.2}{\frac{3.4}{\sqrt{35} } } ) = 0.10

          P(Z \leq \frac{x-16.2}{\frac{3.4}{\sqrt{35} } } ) = 0.10

Now, in the z table the critical value of X which represents the below 10% of the probability area is given as -1.282, that means;

                  \frac{x-16.2}{\frac{3.4}{\sqrt{35} } }  =  -1.282

               x - 16.2  =  -1.282 \times {\frac{3.4}{\sqrt{35} } }

                         x  =  16.2 - 0.74 = <u>15.46 minutes</u>

Hence, the mean oil-change time that would there be a 10% chance of being at or below is 15.46 minutes.

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3 years ago
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