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Natalija [7]
2 years ago
13

During construction the workers need to mix dried bentonite into the concrete as it is being mixed to get the desired consistenc

y. The workers decide to mix 800 mg of Bentonite to be mixed with 198 lb of concrete mix every 8 hours. The information for the concrete shows that 25-75 mg/kg per day is the appropriate mixture consistency.
Are the workers within the desired range? (Note: 1kg = 2.2 lbs)
Mathematics
1 answer:
krek1111 [17]2 years ago
7 0

Answer:

The answer is   A

Step-by-step explanation:

Khan

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4- A manufacturing process produces items whose weights are normally distributed. It is known that 22.57% of all the items produ
galben [10]

Answer:

\\ \mu = 118\;grams\;and\;\sigma=30\;grams

Step-by-step explanation:

We need to use z-scores and a standard normal table to find the values that corresponds to the probabilities given, and then to solve a system of equations to find \\ \mu\;and\;\sigma.

<h3>First Case: items from 100 grams to the mean</h3>

For finding probabilities that corresponds to z-scores, we are going to use here a <u>Standard Normal Table </u><u><em>for cumulative probabilities from the mean </em></u><em>(Standard normal table. Cumulative from the mean (0 to Z), 2020, in Wikipedia) </em>that is, the "probability that a statistic is between 0 (the mean) and Z".

A value of a z-score for the probability P(100<x<mean) = 22.57% = 0.2257 corresponds to a value of z-score = 0.6, that is, the value is 0.6 standard deviations from the mean. Since this value is <em>below the mean</em> ("the items produced weigh between 100 grams up to the mean"), then the z-score is negative.

Then

\\ z = -0.6\;and\;z = \frac{x-\mu}{\sigma}

\\ -0.6 = \frac{100-\mu}{\sigma} (1)

<h3>Second Case: items from the mean up to 190 grams</h3>

We can apply the same procedure as before. A value of a z-score for the probability P(mean<x<190) = 49.18% = 0.4918 corresponds to a value of z-score = 2.4, which is positive since it is after the mean.

Then

\\ z =2.4\;and\; z = \frac{x-\mu}{\sigma}

\\ 2.4 = \frac{190-\mu}{\sigma} (2)

<h3>Solving a system of equations for values of the mean and standard deviation</h3>

Having equations (1) and (2), we can form a system of two equations and two unknowns values:

\\ -0.6 = \frac{100-\mu}{\sigma} (1)

\\ 2.4 = \frac{190-\mu}{\sigma} (2)

Rearranging these two equations:

\\ -0.6*\sigma = 100-\mu (1)

\\ 2.4*\sigma = 190-\mu (2)

To solve this system of equations, we can multiply (1) by -1, and them sum the two resulting equation:

\\ 0.6*\sigma = -100+\mu (1)

\\ 2.4*\sigma = 190-\mu (2)

Summing both equations, we obtain the following equation:

\\ 3.0*\sigma = 90

Then

\\ \sigma = \frac{90}{3.0} = 30

To find the value of the mean, we need to substitute the value obtained for the standard deviation in equation (2):

\\ 2.4*30 = 190-\mu (2)

\\ 2.4*30 - 190 = -\mu

\\ -2.4*30 + 190 = \mu

\\ \mu = 118

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I need help I have 20 min
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Answer:

1- d

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Step-by-step explanation:

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