1.9
9.5% = 0.095
0.095* 20 = 1.9
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.
It would be (all under a big square root) (-7-5)squared + (3+2) squared. This equals to about root 169
Answer:
-7/5
Step-by-step explanation:
The slope of a perpendicular line is the "negative reciprocal" of the slope of the original line. so just switch the numbers and put the top number into a negative.
