Why was 56 and 36 points selected?

Mathematics

1 answer:

AleksAgata [21]3 years ago

6 0

I don’t know , why don’t you tell me why they were selected

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Three potential employees took an aptitude test. Each person took a different version of the test. The scores are reported below

BARSIC [14]

Answer:

Due to the higher Z-score, Demetria should be offered the job.

Step-by-step explanation:

Z-score:

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

In this question:

Whichever applicant had grade with the highest z-score should be offered the job.

Demetria got a score of 85.1; this version has a mean of 61.1 and a standard deviation of 12.

For Demetria, we have $X = 85.1, \mu = 61.1, \sigma = 12$ . So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{85.1 - 61.1}{12}$

$Z = 2$

Vincent got a score of 299.2; this version has a mean of 264 and a standard deviation of 22.

For Vincent, we have $X = 299.2, \mu = 264, \sigma = 22$ .

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{299.2 - 264}{22}$

$Z = 1.6$

Tobias got a score of 7.26; this version has a mean of 7.1 and a standard deviation of 0.4.

For Tobias, we have $X = 7.26, \mu = 7.1, \sigma = 0.4$ .

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{7.26 - 7.1}{0.4}$

$Z = 0.4$

Due to the higher Z-score, Demetria should be offered the job.

3 0

3 years ago

Solve for x<br><img src="https://tex.z-dn.net/?f=4%20%20%7Bx%7D%5E%7B2%7D%20%20%3D%207x%20%2B%202" id="TexFormula1" title="4 {x

Alina [70]

$\bf 4x^2=7x+2\implies 4x^2-7x-2=0\implies (4x+1)(x-2)=0 \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ 4x\cdot x=\boxed{4x^2}~\hfill (1)(-2)=\boxed{-2}\hfill (4x\cdot -2)+(1\cdot x)=\boxed{-7x} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \begin{cases} 4x+1=0\implies 4x=-1\implies &x=-\cfrac{1}{4}\\[1em] x-2=0\implies &x=2 \end{cases}$