Trisha had the following seven distances (in feet) in the triple jump.29, 30, 28, 32, 28, 31, 27What is the approximate mean absolute deviation of her distances? Round the mean and mean absolute deviation to the nearest tenth.0.11.71.510.3

when you find the mean you add all of them together. Then you divide your answer by how many numbers there are that you added. 

So what I did was add 29+30+28+32+29+31+27= 205
then divide 205 by 7 and you get 29.285