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:λ= v X f
v= λ X f
f= λ X v
v= λ / f
Step-by-step explanation:
Answer:
C) Yes, it’s a reflection over line f.
Step-by-step explanation:
It’s a reflection over line f.
Answer:
a. 
Step-by-step explanation:
Since f(x) is the function for the populational density at a certain sidewalk for a 5 mile stretch, a definite integral of that function will yield the total number of people within the integration intervals. If we are interested in the number of people in the whole 5 mile stretch, we must integrate f(x) from x = 0 miles to x = 5 miles:
Therefore, the answer is alternative a.
Answer:
48
f
t
Step-by-step explanation:
