Oftentimes, extremely simple randomized algorithms can achieve reasonably good approximation factors. (a) consider max 3-sat (given a set of 3-clauses, find the assignment that satisfies as many of them as possible). come up with a simple randomized algorithm that will achieve an approximation factor of 7 8 in expectation. that is, if the optimal solution satisfies k clauses, your algorithm should produce an assignment that satisfies at least 7 8 ∗ k clauses in expectation. you may assume that every clause contains exactly 3 distinct variables. (b) given an instance of max 3-sat with n clauses, what is the maximum number of clauses that are guaranteed to be solved in at least one assignment of variables?

if you explain the the question more correctly i might be able to answer this better

step-by-step explanation:

50-38=12

step-by-step explanation:

irst, find the z score of a score of 80. this is given by the formula z = (x-u)/o, where x is the score, u is the mean, and o is the standard deviation. this gives

z = (80-78)/12 = 1/6 = .1667

now, find what percentile a z score of 1/6 correlates to in the normal distribution.

a z score of 1/6 = .1667 is equivalent to the 56.36th percentile of the distribution. so the probability a score is over this is 1-.5636 = .4364

the mean of a sample mean is equal to the original mean. however, the standard deviation of a sample mean is equal to the original standard deviation divided by the square root of n (sample size). so the new standard deviation is 12/sqrt(16) = 3.

the new z score is 80-78/3 = 2/3 = .667. this is equivalent to the 74.86th percentile. so the probability that the mean will be greater than this score is 1-.7486 = .2514.