Answer:
Do the data from this shipment indicate statistical control: No
Step-by-step explanation:
Calculating the mean of the sample, we have;
Mean (x-bar) = sum of individual sample/number of sample
= (0+0+0+0.004+0.008+0.020+0.004+0+0+0.008)/10
= 0.044/10
= 0.0044
Calculating the lower control limit (LCL) using the formula;
LCL= (x-bar) - 3*√(x-bar(1-x-bar))/n
= 0.0044 - 3*√(0.0044(1-0.0044))
= 0.0044- (3*0.0042)
= 0.0044 - 0.01256
= -0.00816 ∠ 0
Calculating the upper control limit (UCL) using the formula;
UCL = (x-bar) + 3*√(x-bar(1-x-bar))/n
= 0.0044 + 3*√(0.0044(1-0.0044))
= 0.0044+ (3*0.0042)
= 0.0044 + 0.01256
=0.01696∠ 0
Do the data from this shipment indicate statistical control: No
Since the value 0.02 from the 6th shipment is greater than the upper control limit (0.01696), we can conclude that the data from this shipment do not indicate statistical control.
Answer:
The answers follow in order, yes, no, and no.
Step-by-step explanation:
<h3>
Answer: 4/13</h3>
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Explanation:
There are A = 13 hearts and B = 4 copies of "7"
The count A+B = 13+4 = 17 may seem like it's the count of hearts or "7" cards, but we're double counting the 7 of hearts. So the true count is actually 17-1 = 16
There are 16 cards we want (either a heart or a "7") out of 52 total. Therefore, the probability is 16/52 = (4*4)/(4*13) = 4/13
Answer:
The first one B
Step by step explanation:
B is (2,4) then do 2+1 which is 3 and then 4 -3 which is 1 and your left with (3,1) which is point B
