The lengths of a lawn mower part are approximately normally distributed with a given mean Mu = 4 in. and standard deviation Sigma = 0.2 in. What percentage of the parts will have lengths between 3.8 in. and 4.2 in.? 34% 68% 95% 99.7%

b) 68%

The percentage of the parts will have lengths between 3.8 in. and 4.2 in

P( 3.8 ≤ x ≤ 4.2) = 0.6826 = 68 %

Step-by-step explanation:

Let 'X' be the normally distributed

mean 'μ'= 4 inches

standard deviation 'σ' = 0.2 inches

Case(i):-

when x₁ = 3.8 inches

Case(ii):-

when  x₂= 3.8 inches

The probability of the parts will have lengths between 3.8 in and 4.2 in

                        = P(Z≤1) - P(Z≤-1)

                        =  0.5 +A(1) -(0.5-A(1)

                       = 2 A(1)

                       = 2×0.3413

                      = 0.6826

Conclusion:-

The percentage of the parts will have lengths between 3.8 in. and 4.2 in

P( 3.8 ≤ x ≤ 4.2) = 0.6826 = 68 %

x = ±sqrt(10)

step-by-step explanation:

5 - 2x² = -15

subtract 5 from each side

5-5 - 2x² = -15-5

-2x^2   = -20

divide by -2 on each side

-2x^2   /-2= -20/-2

x^2 =10

take the square root on each side

sqrt(x^2 )=±sqrt(10)

x = ±sqrt(10)

check:

5 - 2x² = -15

+sqrt(10)

5 - 2(sqrt(10))² = -15

5 -2(10) = -15

5 -20 = -15

-15 = -15

-sqrt(10)

5 - 2(-sqrt(10))² = -15

5 -2(10) = -15

5 -20 = -15

-15 = -15