Difference between 1/5 and 5/5

The weights of players in a basketball league are distributed normally. the distance (d) of each player's weight (w) from the mean of 86 kilograms was computed as follows: d = |w - 86| if d is greater than 7 for approximately 5% of the players, what is the approximate standard deviation of the players' weights? a. 1.75 kilograms b. 3.5 kilograms c. 4.3 kilograms d. 7 kilograms

Add the fractions, and simply the result to lowest terms. 2/3 + 2/7= ?

With finding limits of multi variable !