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:
a. 90
b. 360
Step-by-step explanation:
The time taken by humming bird to flap its wings= 0.08 sec
by flying 7.2 seconds a typical giant hummingbird flap its wings times
= 90 times
Since, A ruby-throated hummingbird can flap its wings 4 times faster than a giant hummingbird, it will flap its wings in every 0.02 sec
so in 7.2 seconds it will flap times = 360 times
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I am not familiar with Laplace transforms, so my explanation probably won't help, but given that for two Laplace transform
and
, then
Given that
and
So you have
From Table of Laplace Transform, you have
and hence
So you have
.
Hope this helps...