If I'm reading this correctly, it is 2*3*n*n = 6n^2
6n^2 = 54 Divide by 6
n^2 = 54/6
n^2 = 9 Take the square root of both sides.
sqrt(n^2) = sqrt(9)
n = +/- 3
n = + 3 or n = -3. Both will solve the equation above.
Edit
2 * 3 * n * n
Let n = 3
2 * 3 * 3 * 3 = 6 * 9 = 54
The other value is n = - 2
2 * 3 * n * n
2 * 3 * (-3) * (-3) A minus times a minus is a plus.
2 * 3 * 9 = 6 * 9 = 54
both + 3 and - 3 for n will give 54.
Answer:
thus the probability that a part was received from supplier Z , given that is defective is 5/6 (83.33%)
Step-by-step explanation:
denoting A= a piece is defective , Bi = a piece is defective from the i-th supplier and Ci= choosing a piece from the the i-th supplier
then
P(A)= ∑ P(Bi)*P(C) with i from 1 to 3
P(A)= ∑ 5/100 * 24/100 + 10/100 * 36/100 + 6/100 * 40/100 = 9/125
from the theorem of Bayes
P(Cz/A)= P(Cz∩A)/P(A)
where
P(Cz/A) = probability of choosing a piece from Z , given that a defective part was obtained
P(Cz∩A)= probability of choosing a piece from Z that is defective = P(Bz) = 6/100
therefore
P(Cz/A)= P(Cz∩A)/P(A) = P(Bz)/P(A)= 6/100/(9/125) = 5/6 (83.33%)
thus the probability that a part was received from supplier Z , given that is defective is 5/6 (83.33%)
7-2c= c - 2 this is the answer
Answer:
5/8
Step-by-step explanation:
