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%)
6j + 1/3n = 134...multiply by 3....18j + n = 402
1/3j + n = 31....multiply by 3....j + 3n = 93
the multiplying by 3 is optional...I just did it because it is easier to work with equations when there are no fractions.
18j + n = 402....multiply by -3
j + 3n = 93
----------------
-54j - 3n = - 1206 (result of multiplying by -3)
j + 3n = 93
----------------add
-53j = - 1113
j = -1113/-53
j = 21
j + 3n = 93
21 + 3n = 93
3n = 93 - 21
3n = 72
n = 72/3
n = 24
so Jasons collection (j) consists of 21 books and Nathans collection (n) consists of 24 books
Answer:
C. Incomplete sentences
Step-by-step explanation:
Answer:
look down
Step-by-step explanation:
OK so u take something like 9/3 and then ask how many times the 3 goes into the 9 and that is your whole number. whatever is left is your top of the fraction and the bottom stays the same hope this helped
