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.
Based on the financial details given, the total amount of current liabilities on xyz company's 12/31/x1 classified balance sheet is $22,600.
<h3>What are the total current liabilities?</h3>
The current liabilities can be found by the formula:
= Accounts Payable + Salaries Payable + Interest Payable + Utilities Payable
As the above are all the current liabilities to XYZ company, the current liabilities are:
= 11,300 + 2,500 + 1,900 + 6,900
= $22,600
Find out more on balance sheets at brainly.com/question/1113933
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4Th option is correct ✔️ ✔️
7 (x - 20)
use distribution method
7x - 140