Answer:
Step-by-step explanation:
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%)
Answer:
0.82644628099
Step-by-step explanation:
The answer is 75 and here's why.
Let's figure out the amount of remaining votes to prove this.
250 total votes - 150 votes for candidate a = 100 votes.
minus ↑ sign
Candidate B received 25% of the votes, so let's find<u> 25% of 100</u>
<u>100 * 0.25 = 25 votes</u>
Candidate B only got 25 votes (that's kinda sad, poor guy)
<u>100 - 25 votes = the # of votes candidate C got </u>
Candidate C got 75 votes!
Step-by-step explanation:
EF = 4x - 15
FG = 3x - 7
EG = EF + FG = 20
so,
4x - 15 + 3x - 7 = 20
7x - 22 = 20
7x = 42
x = 6
EF = 4×6 - 15 = 9
FG = 3×6 - 7 = 11
