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Remember
(x^m)/(x^n)=x^(m-n)
(q^(33/4))/(q^8)=q^((33/4)-8)
what is 33/4-8?
33/4-8=33/4-32/4=1/4
result is
q^1/4 or![\sqrt[4]{q}](https://tex.z-dn.net/?f=%20%5Csqrt%5B4%5D%7Bq%7D%20)
answer is
(x^m)/(x^n)=x^(m-n)
(q^(33/4))/(q^8)=q^((33/4)-8)
what is 33/4-8?
33/4-8=33/4-32/4=1/4
result is
q^1/4 or
answer is
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 =
B) Probability that the selected employee has been with the firm between 5 and 10 years
=
C) Probability that the selected employee has been with the firm more than 10 years
=
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)