C is the answer radical 77/22
Let p be 0.831 denote the percentage of defective welds and q be 0.169 denote the percentage of non-defective welds.
Using the binomial distribution, we want all three to be defective.


If the number to the right to the number you are rounding higher than five add one to the number on the left all the numbers behind the eight turn into zeros the answer is 900,000