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Arisa [49]
3 years ago
5

Nadia is asked to find a number such that the difference between five times the number and three less than twice the number is e

qual to eighteen. Her work and answer are shown below. 5n−(3−2n)5n−3+2n7n−37nn=====181818213 Is Nadia correct? If not, identify and correct her error.
Mathematics
2 answers:
kolbaska11 [484]3 years ago
6 0

Answer:

Nadia is incorrect, please check explanations for the correction to her work

Step-by-step explanation:

Here, we want to validate the wok of Nadia

Let the number be n

5 times the number is 5 * n = 5n

Difference between 5n and 3 less than twice the number

3 less than twice the number is 2n-3

So it should be 2n-3 ;

The solution is thus;

5n - (2n-3) = 18

5n - 2n + 3 = 18

3n + 3 = 18

3n = 18-3

3n = 15

n = 15/3

n = 5

Nataliya [291]3 years ago
4 0

Answer:

Nadia is incorrect; she did not translate the given information into an equation properly. The first equation should be 5n−(2n−3)=18.

Step-by-step explanation:

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X + 2(y-6) = 0
Karo-lina-s [1.5K]
2x - z = 4
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If u want me to solve them it’s
1. X = -2y + 12
2. x = 10 - 4/3y | for y, y = 15/2 - 3/4x
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2 years ago
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

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

As we say before, the z-score "tells us" the distance from the population mean, and in this case this value equals zero:  

\\ x = \mu

Therefore

\\ z = 0

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

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