Ummm is this a full question?
Based on the given table above, the correct answer would be option D. 2.18%. So how did we get this result. Looking at the table, the number of adults experiencing nausea is 24 and the total number of trial members is 1100. So given these values, the p<span>robability of getting nausea (as a percentage) is
(24/1100)*100. (0.0218)*100, so we get 2.18. Which gives the probability of 2.18%. Hope this answer helps.</span>
16-A 18-B 19-D (not 100% sure on 16)
