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.
First let's get rid of the parenthesis:
12 - 4x + 3x = 4 + 10 + 2x
Now group the x's on the left and the numbers on the right
-4x + 3x - 2x = 4 + 10 - 12
...and simplify
-3x = 2
divide by -3 to isolate x
x = -2/3
The answer is the graph b.
Sin x = 0.5
sin x = 1/2
sin x = opp/hyp
therefore the ratio of opp/hyp = 1/2, (opp = 1, hyp = 2)
Find the opp side
1² + x² = 2²
1 + x² = 4
x² = 4 -1
x² = 3
x =√3
The opp side is √3
cos x = opp/hyp = √3/2 = 0.87 (round to the nearest hundredths)
