Let
....(1)
Multiply both sides of equation (1) by 10.
.... (2)
Multiply both sides of equation (1) by 100.
... (3)
Subtract equation (2) from equation (3)



Hence, the required fraction form is
.... Answer.
Answer:
The standard deviation of the number of rushing yards for the running backs that season is 350.
Step-by-step explanation:
Consider the provided information.
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.
Here it is given that mean is 790 and 1637 is 2.42 standard deviations above the mean.
Use the formula: 
Here z is 2.42 and μ is 790, substitute the respective values as shown.



Hence, the standard deviation of the number of rushing yards for the running backs that season is 350.
Lets take Gregs weight as “x”. This means that Justins weight is x-15, and x/2 = (x-15)-75.
If we take that last equation, lets combine like terms:
x/2 = x - 15 - 75
x/2 = x - 90
Now multiply both sides by 2 to get rid of the fraction
x = 2x - 180
Subtract 2x from both sides
x - 2x = -180
-x = -180
x = 180 — this is Gregs weight
Justins weight is x-15, so 180-15, which is 165 pounds. Hope this helped.