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levacccp [35]
3 years ago
5

Tom compared \dfrac{2}{12} 12 2 ​ start fraction, 2, divided by, 12, end fraction and \dfrac{7}{6} 6 7 ​ start fraction, 7, divi

ded by, 6, end fraction by first comparing each fraction to \dfrac{1}{2} 2 1 ​ start fraction, 1, divided by, 2, end fraction and 111. Step 1: \dfrac{2}{12} 12 2 ​ start fraction, 2, divided by, 12, end fraction is less than \dfrac{1}{2} 2 1 ​ start fraction, 1, divided by, 2, end fraction. Step 2: \dfrac{7}{6} 6 7 ​ start fraction, 7, divided by, 6, end fraction is less than 111 but more than \dfrac{1}{2} 2 1 ​ start fraction, 1, divided by, 2, end fraction. Step 3: So, \dfrac{2}{12} < \dfrac{7}{6} 12 2 ​ < 6 7 ​ start fraction, 2, divided by, 12, end fraction, is less than, start fraction, 7, divided by, 6, end fraction In which step did Tom make his first mistake?
Mathematics
2 answers:
Alex Ar [27]3 years ago
7 0
It’s 3 so make sure to put that as the answer ok your welcome
soldier1979 [14.2K]3 years ago
6 0

B hope that helps! Btw, it's on khan so put B and it should work!

                                                                                                       

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The results of a common standardized test used in psychology research is designed so that the population mean is 155 and the sta
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Answer:

The value <em>155</em> is zero standard deviations from the [population] mean, because \\ x = \mu, and therefore \\ z = 0.

Step-by-step explanation:

The key concept we need to manage here is the z-scores (or standardized values), and we can obtain a z-score using the next formula:

\\ z = \frac{x - \mu}{\sigma} [1]

Where

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  • x is the <em>raw score</em>: an observation from the normally distributed data that we want <em>standardize</em> using [1].
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  • \\ \sigma is the <em>population standard deviation</em>.

Carefully looking at [1], we can interpret it as <em>the distance from the mean of a raw value in standard deviations units. </em>When the z-score is <em>negative </em>indicates that the raw score, <em>x</em>, is <em>below</em> the population mean, \\ \mu. Conversely, a <em>positive</em> z-score is telling us that <em>x</em> is <em>above</em> the population mean. A z-score is also fundamental when determining probabilities using the <em>standard normal distribution</em>.

For example, think about a z-score = 1. In this case, the raw score is, after being standardized using [1], <em>one standard deviation above</em> from the population mean. A z-score = -1 is also one standard deviation from the mean but <em>below</em> it.

These standardized values have always the same probability in the <em>standard normal distribution</em>, and this is the advantage of using it for calculating probabilities for normally distributed data.

A subject earns a score of 155. How many standard deviations from the mean is the value 155?

From the question, we know that:

  • x = 155.
  • \\ \mu = 155.
  • \\ \sigma = 50.

Having into account all the previous information, we can say that the raw score, <em>x = 155</em>, is <u><em>zero standard deviations units from the mean.</em></u> <u><em>The subject   earned a score that equals the population mean.</em></u> Then, using [1]:

\\ z = \frac{x - \mu}{\sigma}

\\ z = \frac{155 - 155}{50}

\\ z = \frac{0}{50}

\\ z = 0

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\\ x = \mu

Therefore

\\ z = 0

So, the value 155 is zero standard deviations <em>from the [population] mean</em>.

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