Answer:
Let x = number of units produced per hour before the line was modified , y = number of units produced per hour after the line was modified.
The null and alternative hypotheses are :
H0: μbefore = μafter
H1: μbefore ≠ μafter
Step 1: . Calculate the difference (D = y − x) between the two observations on each pair.
See attachment for (D=y-x)
step 2: Calculate ∑D and ∑D2
From table we get ∑D = -10 & ∑D2 = 72
Step 3: Put all the value in test statistic "t"
t = D/√nD^2-D^2/n - 1
t =-10/√6.72-(-10)^2/6-1
t= -10√66.4
= -10/8.1486
= -1.227
Step 4: Compare tcal and ttab
At α = 0.05
t0.05 for 5 d.f. = 3.1634 (two tail test)
Hence | tcal | < t0.05
So, we accept the hypothesis.
So we conclude that the modified (after) layout has not increased worker productivity at 5% level of significance.
Answer: 5 cm
Step-by-step explanation:
B = 
Answer:
Your answer is correct
Step-by-step explanation:
Indeed subtracting 5 from both sides would be n/-3> -1
Multiplying both sides by -3 would give you n < 3 (obviously after reversing the inequality sign)
Answer:
0.001709
Step-by-step explanation:
The probability of getting 9 tails then 1 head is:
(½)⁹ (½)¹ = (½)¹⁰
The probability of getting 10 tails then 1 head is:
(½)¹⁰ (½)¹ = (½)¹¹
The probability of getting 11 tails then 1 head is:
(½)¹¹ (½)¹ = (½)¹²
The total probability is:
(½)¹⁰ + (½)¹¹ + (½)¹²
0.001709
42 cm because if you divide 210 and 1260 it will make 6 so u just multiply it by 7 and u get 42
