The Acme Company manufactures widgets. The distribution of widget weights is bell-shaped. The widget weights have a mean of 50 ounces and a standard deviation of 10 ounces. Use the Empirical Rule, what percentage of the widget weights lie between 43 and 87 ounces?

The probability of the widget weights lie between 43 and 87 ounces

P(43 ≤ x ≤ 87 ) = 0.7605

The percentage of the widget weights lie between 43 and 87 ounces

P(43 ≤ x ≤ 87 ) = 76 %

Step-by-step explanation:

Step(i):-

Given mean of the Population = 50ounces

Given standard deviation of the Population = 10 ounces

Let 'X' be the random variable in Normal distribution

Given X = 43

Given X = 87

Step(ii):-

The probability of the widget weights lie between 43 and 87 ounces

                       = |A(3.7) + A(-0.7)|  

                       =  |A(3.7) + A(0.7)|   (∵A(-z) = A(z))

                      = 0.4994 + 0.2611

                     =  0.7605

Final -

The probability of the widget weights lie between 43 and 87 ounces

P(43 ≤ x ≤ 87 ) = 0.7605

The percentage of the widget weights lie between 43 and 87 ounces

P(43 ≤ x ≤ 87 ) = 76 %

answer is 8x+24

area of vanessa's patio = length × width

given, length = (3x² + 5x + 10)

width = (x² - 3x - 1)

area of vanessa's patio = (3x² + 5x + 10)(x² - 3x - 1)

= 3x²(x² - 3x - 1) + 5x(x² - 3x - 1) + 10(x² - 3x - 1)

= 3x⁴ - 9x³ - 3x² + 5x³ - 15x² - 5x + 10x² - 30x - 10

= 3x⁴ - 4x³ - 8x² - 35x - 10

hence, expression of vanessa's patio = (3x⁴ - 4x³ - 8x² - 35x - 10)