Suppose that 95% of the fasteners pass the initial inspection. Of those that fail the initial inspection, 20% are defective. Of the fasteners sent to the recrimping operation, 40% cannot be corrected ad are scrapped; the are corrected by the recrimping and then pass inspection.

A. What proportion of the fasteners that fail the initial inspection pass the second inspection(after the recrimping operation)?
B. What proportion of fasteners pass inspection?
C. Given that a fastener passes inspection, what is the probability that it passed the initial inspection and did not have to go through the recrimping operation?

Answer

Step-by-step explanation:

Let x be the fasteners

95% of x passed, and 5% of x failed

20% of the 5% which failed are defective, remaining the 80% of the failed 5%.

Now 40% of this 80% of the 5% which failed are scrapped, meaning only the remaining 60% passed second inspection.

Thus

a). The proportion ;

60% of 80% of 5% of x

mathematically

= 60/100 × 80/100 × 5/100

= 0.024 or 2.4%

b) initially 95% passed and later on 2.4% pass again. Therefore total that passed = 95%+2.4%=97.4%

c) probability that it passed on first inspection is 95/100

Percentage that went through recrimping = 80% of 5% = 80/100 × 5/100 = 1/25 or 0.04 or 4%

Thus probability of passing and not going through recrimping

= 1 - (1/25) = 0.96 or 96%

395.84

step-by-step explanation:

2πrh+2πr²

2 (pie)(radius)(height)+2(pie)(radius)^2

answer: 59

step-by-step explanation: