Answer:
(a). 72.9%.
(b). 13.6 hr.
Step-by-step explanation:
So, we are given the following data or parameters or information which is going to assist us in solving this question/problem;
=> "A welder produces 7 welded assemblies during the first day on a new job, and the seventh assembly takes 45 minutes (unit time). "
=> The worker produces 10 welded assemblies on the second day, and the 10th assembly on the second day takes 30 minutes"
So, we will be making use of the Crawford learning curve model.
T(7) + 10 = T (17) = 30 min.
T(7) = T1(7)^b = 45.
T(17 ) = T1(17)^b = 30.
(T1) = 45/7^b = 30/17^b.
45/30 = 7^b/17^b = (7/17)^b.
1.5 = (0.41177)^b.
ln 1.5 = b ln 0.41177.
0.40547 = -0.8873 b.
b = - 0.45696.
=> 2^ -0.45696 = 0.7285.
= 72.9%.
(b). T1= 45/7^ - 045696 = 109.5 hr.
V(TT)(17) = 109.5 {(17.51^ - 0.45696 – 0.51^ - 0.45696) / (1 - 0.45696)} .
V(TT) (17) = 109.5 {(4.7317 - 0.6863) / 0.54304} .
= 815.7 min .
= 13.595 hr.
Answer:
yes 30 is the answer
Step-by-step explanation:
Answer:
E. P(W | H)
Step-by-step explanation:
What each of these probabilities mean:
P(H): Probability of the game being at home
P(W): Probability of the game being a win.
P(H and W): Probability of the game being at home and being a win.
P(H|W): Probability of a win being at home.
P(W|H): Probability of winning a home game.
Which of the following probabilities do you need to find in order to determine the proportion of at-home games that were wins?
This is the probability of winning a home game. So the answer is:
E. P(W | H)
Answer:
-34.5≥ x
Step-by-step explanation:
- Simply because the values start from -34.5 and then decrease
