A vat of water and a weighing scale
It is 45 degrees
that is your answer
Answer:D
Step-by-step explanation:
Answer:
The best choice would be hiring a random employee from company A
Step-by-step explanation:
<em>Supposing that the performance rating of employees follow approximately a normal distribution on both companies</em>, 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
L = 5W - 10
5W = L + 10
W = 1/5(L + 10)
Area = L x W
240 = L(1/5(L + 10)
L^2 + 10L = 240(5) = 1200
L^2 + 10L - 1,200 = 0