Without properly working spark plugs, a vehicle will not run. For a specific vehicle, the spark plugs are supposed to have a gap between 3.9mm and 4.3mm. Any spark plugs with gaps larger or smaller than this will fail inspection and be discarded. At the factory, the machine that sets the gap follows a normal distribution with a mean of 4.1mm and standard deviation of 0.075mm. What is the probability that a randomly selected spark plug passes inspection

The probability that the spark plugs are supposed to have a gap between 3.9mm and 4.3mm.

P(3.9≤X≤4.3)  = 0.9922

Step-by-step explanation:

Step(i):-

Given that the mean of the Normal distribution = 4.1mm

Given that the standard deviation of the Normal distribution = 0.0075mm

Let 'X' be the random variable in a normal distribution

Given that X₁ = 3.9mm

Given that X₂ = 4.3mm

Step(ii):-

The probability that the spark plugs are supposed to have a gap between 3.9mm and 4.3mm.

P(3.9≤X≤4.3) = P(-2.666≤Z≤2.666)

                     = P(Z≤2.666)-P(Z≤-2.666)

                    = 0.5 +A(2.666) - (0.5-A(2.666)

                   = 2 × A(2.666)

                 = 2×0.4961

                 = 0.9922

Final -

The probability that the spark plugs are supposed to have a gap between 3.9mm and 4.3mm.

P(3.9≤X≤4.3)  = 0.9922

2

Step-by-step explanation:because you are trying to find A-C so that means (9x+8)+(4x-2)=line AC Line Ac also is 3x+26. You set them to equal, getting 13x+6=3x+26 subtract 3x both sides, 26=10x+6    Subtract 6 both sides, 20=10x  divide by 10 and you get x=2