1.) set up a w_k_u chart
2.) in the w column write in on top and mi in the bottom like this
w_K_U
in_
mi_
then in the k column write 2 for in and 15 for mi
W_K_U
in_2_
mi_15_
then in the U column, you write 7 for in and x for mi
W_K_U
in_2_7
mi_15_x
then cross multiply: 2x = 15(7)
multiply 15 times 7 which is 105
2x=105
divide 2x by 2
x=105
answer 105
Median, upper quartile, and maximum are all incorrect
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.
66.7 %
convert to decimal the multiply by 100
Step-by-step explanation:
Slope-intercept form: y = mx + c
y = x + 3 is already in this form.
3x + 3y = -9,
3y = -3x - 9
y = -x - 3.
