Instruction: you must show all of your work bysl setting up the problem using combinatorics and simplifying completely. question: there are 5 red marbles and 9 blue marbles in a bag. suppose two of the marbles are cracked. how many ways can four marbles be selected so that either no cracked marbles are selected or just one cracked marble is selected?

There are 14 marbles in the bag, and we either want to get 0 or 1 cracked one (out of 2 cracked marbles) in 4 draws. Since there are 12 uncracked marbles and 2 cracked ones:
To get 0 cracked marbles, there are 12C4 ways (to draw 4 marbles out of the 12 uncracked ones only). 12C4 = 495 ways
To get 1 cracked marble, we need 3 to be from the 12 uncracked ones, and 1 from the 2 cracked ones. This means there are 12C3 x 2C1 = 440 ways
So the total number of ways is 495 + 440 = 935 ways.

You need to divide 210 square inches by 35 inches and that’s going to give you 6 inches and that’s the length of the hypothenuse of the right triangle i hope that this you : )