The times for the mile run of a large group of male college students are approximately normal with mean 7.09 minutes and standard deviation 0.77 minutes. use the 68-95-99.7 rule to answer the following questions. (start by making a sketch of the density curve you can use to mark areas on.) (a) what range of times covers the middle 68% of this distribution? (b) what percent of these men run a mile in more than 9.4 minutes? (a) from 6.32 minutes to 7.86 minutes (b) 13 %

c. y = 1/4 x + 5

step-by-step explanation:

first, put the equation of the line on the graph into slope-intercept form

(y = mx + b):   y = -4x + 8

second, a perpendicular line has to have "an opposite/reciprocal slope" to the original slope. so, find the "m" value in your original equation (it's -4, right? )

original slope: m = -4   perpendicular slope: change the sign to its opposite and find the reciprocal ("flip" the number)

new slope is now positive (b/c the original slope was negative 4), and it's 1/4 (not the original slope of 4 or 4/1)

answer:   x = 3 + √11     and     x = 3 - √11

step-by-step explanation:

x² - 6x + 9 = 11

(x - 3)² = 11

x - 3 = ±√11

    x = 3 ± √11

x = 3 + √11     and     x = 3 - √11