answer:To find the complement of set A with respect to universal set U , we need to identify all the elements that are in U but not in A . The universal set U consists of the elements {1, 2, 3, 4, 5} . The set A contains the elements {2, 3, 4} . The complement of A , complement_U A , will therefore include all the elements from U that are not present in A . We list out these elements: complement_U A = {1, 5} Therefore, the solution to the problem is: boxed{complement_U A = {1, 5}}

answer:**Analysis** This question mainly examines the application of systematic sampling, which is a common type of question in college entrance examinations and is considered a basic question. **Solution** Given the information, the interval distance is dfrac{56}{4} = 14. Therefore, the number of the remaining student in the sample is 19. Hence, the answer is boxed{19}.

answer:Let's denote the total number of balls in the bag as ( T ). We know that there are 47 red balls and 3 purple balls, so there are ( 47 + 3 = 50 ) balls that are either red or purple. The probability that a ball chosen at random will be neither red nor purple is given as 0.5 (or 50%). This means that the number of balls that are neither red nor purple must be equal to half of the total number of balls in the bag. We also know the number of balls that are neither red nor purple: 10 white, 30 green, and 10 yellow, which adds up to ( 10 + 30 + 10 = 50 ) balls. So, if the probability of choosing a ball that is neither red nor purple is 0.5, then the number of balls that are neither red nor purple (50) must be half of the total number of balls in the bag. Therefore, we can set up the equation: [ frac{50}{T} = 0.5 ] To find ( T ), we multiply both sides by ( T ) and then divide both sides by 0.5: [ 50 = 0.5T ] [ T = frac{50}{0.5} ] [ T = 100 ] So, there are boxed{100} balls in total in the bag.

answer:1. **Identify Similar Triangles**: We need to determine the similarity between two triangles in our problem. We consider triangles KOA and KEO. Note that angle angle EOK = angle OAK. **Angles**: - angle EOK: This is a common angle for both triangles KOA and KEO. - angle OAK: This angle is shared between these triangles because angle OAK and angle AKE are corresponding angles created by the transversal intersecting parallel lines and the circle. 2. **Establish the Proportion**: Given the triangles KOA and KEO are similar by the AA (Angle-Angle) criterion of similarity: [ angle EOK equiv angle OAK ] We can write the similarity relation as follows: [ frac{KE}{OK} = frac{OK}{AK} ] 3. **Relate the Segments**: From the similarity of triangles, squaring both sides of the proportion: [ (OK)^2 = KE times AK ] 4. **Apply the Tangent-Secant Theorem**: According to the Tangent-Secant Theorem, which states that for a point K outside a circle, the square of the length of the tangent segment is equal to the product of the whole secant segment (from the external point to the farthest intersection point on the secant with the circle) and its external segment (from the external point to the nearest intersection point on the secant). [ KE times AK = (KB)^2 ] 5. **Conclude Equality**: Since we have: [ OK^2 = KE times AK ] and [ KE times AK = (KB)^2 ] it logically follows: [ OK = KB ] # Conclusion: Therefore, we have shown that: [ boxed{OK = KB} ]