For a recent season in college football, the total number of rushing yards for that season is recorded for each running back. The mean number of rushing yards for the running backs that season is 790 yards. One running back had 1,637 rushing yards for the season, which is 2.42 standard deviations above the mean number of rushing yards. What is the standard deviation of the number of rushing yards for the running backs that season? (A) 250 yards

(B) 300 yards

(C) 350 yards

(D) 400 yards

(E) 450 yards

(C) 350 yards

Step-by-step explanation:

From available information

Mean = 790

1 running back = 1637 yards

Standard deviation above mean = 2.42

Using formula

z =( x - mean)/sd

2.42 = (1637-790)/sd

2.42 = 847/sd

We cross multiply

2.42sd = 847

Divide through by 2.42

Sd = 847/2.42

Sd = standard deviation = 350

So our answer is 350 and therefore option c.

answer: domain and range mean all possible values of x and y that could be substituted to the inequality. a solution means all possible values that make the inequality statement true. so while the range is still all real numbers, the solution is the shaded area, for which y will always be greater than the line of the function.

answer: options

step-by-step explanation: