Kathleen missed 15 out out of her last 20 free throws. consendering this data, how many of her next 12 free-throw attempts would you expect kathleen to miss

Best answer that i found on yahoo ok   set :   x= length of 2nd piece in cm  x- 3= length of 1st piece in cm  x-4= length of 3rd piece in cm  then since you know they all came from the 20 cm ribbon you add them together and set them equal to 20x+ x-3 +x-4 = 20  3x - 7= 20  3x= 27  x= 9  the shortest piece is the third piece which is x- 4  x-4  9-4 = 5  the length of the smallest piece is 5 cm.  the other lengths are  9 cm  6 cm

3cm and 6 and 9 cm that a possibility of that question