The average performance rating of the employees at company a is 6.5 on a scale from 1 to 10 (10 being highest). the standard deviation is 2.1 the average performance rating of the employees at company c is 7.4. company c has a standard deviation of 6.8. it would be better to hire a random employee from company a.

The best choice would be hiring a random employee from company A

Step-by-step explanation:

Supposing that the performance rating of employees follow approximately a normal distribution on both companies, we are interested in finding what percentage of employees of each company have a performance rating greater than 5.5 (which is the mean of the scale), when we measure them in terms of z-scores.

In order to do that we standardize the scores of both companies with respect to the mean 5.5 of ratings

The z-value corresponding to company A is

where

= mean of company A

= 5.5 (average of rating between 1 and 10)

s = standard deviation of company A

We do the same for company C

This means that 27.49% of employees of company C have a performance rating > 5.5, whereas 71.42% of employees of company B have a  performance rating > 5.5.

So, the best choice would be hiring a random employee from company A

341

step-by-step explanation:

all you have to do is 9x31 (which equals 279) and then 20x31 (which equals 620). then you have to subtract. 620-279=341.

6mm^3 or .6cm^3

      216

      /   \

  36     6

  /   \     /

6     6   6

factor trees are nice.