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PtichkaEL [24]
3 years ago
10

a high driver stands on a cliff 150 feet above sea level. he dives into the sea and goes 15 feet below the water. what is the di

stance from where he stood on the cliff to where he dived?
Mathematics
2 answers:
MrRissso [65]3 years ago
6 0
The high diver dived 165 feet
natulia [17]3 years ago
3 0
Start with the 150 and then add the 15, 165
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Il me semble que le triangle MNP devrait ressembler à celui que j’ai attaché.

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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
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\\ \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}

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<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)

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Rearranging these two equations:

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

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

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\\ 3.0*\sigma = 90

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

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

\\ 2.4*30 - 190 = -\mu

\\ -2.4*30 + 190 = \mu

\\ \mu = 118

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