answer:1. **Total Outcomes**: Since only two dice are re-rolled, there are (6^2 = 36) possible sets of dice rolls for these two dice. 2. **Successful Outcomes**: - **Case 1**: At least one of the re-rolled dice matches the three set aside. Each die has one specific value it needs to show (the same as the set-aside three), which has a (1/6) chance of occurring. Thus, the probability either of the two dice matches the set aside three is calculated as: [1 - P(text{neither matches}) = 1 - left(frac{5}{6} times frac{5}{6}right) = 1 - frac{25}{36} = frac{11}{36}] - **Case 2**: The two re-rolled dice match each other (but do not match the three set aside). There are 5 options for this (since they can't match the initial three-of-a-kind), so there are (frac{5}{36}) chances for this. 3. **Total Successful Outcomes**: [text{Probability} = frac{11}{36} + frac{5}{36} = frac{16}{36} = frac{4}{9}] Conclusion: Thus, the probability that at least three of the five dice show the same value after re-rolling the two dice is (boxed{frac{4}{9}}).

answer:Given a > 0 > b > -a, and c < d < 0, we have: (1) ad > bc is not true; (2) frac{a}{b} + frac{b}{c} < 0 is true; (3) a - c > b - d is true; (4) a(d - c) > b(d - c) is true; Therefore, the answer is: (2), (3), (4). From the given conditions a > 0 > b > -a, and c < d < 0, by analyzing the validity of the inequalities in the four options based on the properties of inequalities, we can find the answer. This problem tests the knowledge of inequality relations and inequalities, where mastering the basic properties of inequalities is key to solving the problem. Thus, the correct statements are boxed{(2), (3), (4)}.

answer:Given that a_3a_7 = a_5^2 = 64, and since a_n > 0, the value of a_5 is 8. Therefore, the correct option is boxed{D}.

answer:Since {a, b, c} = {0, 1, 3}, the elements a, b, and c have the possible values: - When a = 0, there are two scenarios for b and c: if b = 1 and c = 3 or if b = 3 and c = 1. Both scenarios do not meet the given conditions since two of the original three statements would be true. - When a = 1, the possibilities are: b = 0 and c = 3 or b = 3 and c = 0. Again, neither of these scenarios meet the conditions considering two statements would be true. - When a = 3, there are two scenarios: if b = 0 and c = 1 or if b = 1 and c = 0. In this case, if a = 3, then statement (1) a neq 3 is incorrect, statement (2) b = 3 is incorrect, but statement (3) c neq 0 is correct. This meets the condition of only one statement being correct. Therefore, the values of a, b, and c are 3, 0, and 1 respectively. Plugging these into 100a + 10b + c gives: 100a + 10b + c = 100 times 3 + 10 times 0 + 1 = 300 + 1 = boxed{301}