2x/3-6=4. 2x/3=4+6. 2x/3=10. 2x=10*3. 2x=30. X=30/2. x=15.
Answer:
Step-by-step explanation:
Given that a small manufacturing firm has 250 employees. Fifty have been employed for less than 5 years and 125 have been with the company for over 10 years. So remaining 75 are between 5 and 10 years.
Suppose that one employee is selected at random from a list of the employees
A) Probability that the selected employee has been with the firm less than 5 years = ![\frac{50}{250 } \\= 0.20](https://tex.z-dn.net/?f=%5Cfrac%7B50%7D%7B250%20%7D%20%5C%5C%3D%200.20)
B) Probability that the selected employee has been with the firm between 5 and 10 years
= ![\frac{75}{250 } \\= 0.30](https://tex.z-dn.net/?f=%5Cfrac%7B75%7D%7B250%20%7D%20%5C%5C%3D%200.30)
C) Probability that the selected employee has been with the firm more than 10 years
= ![\frac{125}{250} =0.50](https://tex.z-dn.net/?f=%5Cfrac%7B125%7D%7B250%7D%20%3D0.50)
a) P(A) = 0.2
P(C) = 0.5
P(A or B) = 0.2+0.3 = 0.5
P(A and C) = 0 (since A and C are disjoint)
The probability that the monthly payment is more than $1000 will be found as follows;
The payment is normally distributed, thus the z-score will be given by:
Z-score=(x-Mean)/(SD)
Mean=$982
SD=$180
Thus;
Z-score=(1000-982)/180=0.1
The probability associated with a z-score of 0.1 is 0.5398
Thus the probability that the monthly payment is more than $1000 will be:
P(x<1000)=1-0.5398=0.4602=46.02%