Answer:
I also don't know the answer to this question
Answer:
Probability that this whole shipment will be accepted is 0.7324.
Step-by-step explanation:
We are given that a company purchases shipments of machine components and uses this acceptance sampling plan: Randomly select and test 30 components and accept the whole batch if there are fewer than 3 defectives.
The above situation can be represented through Binomial distribution;

where, n = number of trials (samples) taken = 30 components
r = number of success = fewer than 3
p = probability of success which in our question is % rate
of defects, i.e; 6%
<em>LET X = Number of defective components</em>
So, it means X ~ 
Now, Probability that this whole shipment will be accepted only when there are fewer than 3 defectives = P(X < 3)
P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)
=
=
= 0.7324
Therefore, probability that this whole shipment will be accepted is 0.7324.
Answer:
d = 17.58
Step-by-step explanation:
d = √(x2 - x1)^2 + (y2 - y1)^2 + (z2 - z1)^2
P1(0, 0, 0) P2(8, 7, 14)
Substitute the values in the above equation and you will get
d = √(8 - 0)^2 + (7 - 0)^2 + (14 - 0)^2
= √(64 + 49 + 196)
= √309
= 17.58
Answer:
1/2
I believe it would be 1/2
Answer:
it is b
Step-by-step explanation: