Answer:
None of the Above
Explanation: Candidate key is a field in an RDBMS which independently identifies each unique record in a row. It has distinct tuples and no redundant attributes. Here, in the given table, there are no such rows or tuples. Hence, the answer is None of the Above.
R Consider an instance of relation R below A B C D E 5 5 5 5 5 5 5 5 5 4 5 2 4 5 2 4 1 4 5 5 5 4 2. 5 4 4 1 2 4 2 What would possibly be a candidate key of relation R? (2 points) Select one: AE BC BD CD All of the above. None of the above.
Consider an instance of relation R below 20 A B С D E 5 5 5 5 5 5 5 5 5 4 5 2 4 2 4 1 4 ي ي اي 5 5 5 4 2 4 4 1 4 2 N What would possibly be a candidate key of relation R? (2 Points) Select one: O AE BC BD CD All of the above. None of the above. O
Consider an instance of relation R below R A B С DE 5 5 5 5 5 5 5 5 5 4 5 2 4 2 4 1 4 o con 5 5 5 4 4 N 4 1 4 2 N What would possibly be a candidate key of relation R? (2 Points) Select one: O AE BC BD CD All of the above. None of the above.
Let R(A,B,C,D,E) be a relation with FDs F = {AB-C, CD-E, E-B, CE-A} Consider an instance of this relation that only contains the tuple (1, 1, 2, 2, 3). Which of the following tuples can be inserted into this relation without violating the FD's? (2 points) Select one: O (0, 1, 2, 4,3) (1,1,2,2,4) (1.2.2, 2, 3) o (1,1,3,2,3) All of the above can be inserted. None of the above can be inserted.
Let R(A, B, C, D, E) be a relation wit FDs F = {AB->C, CD->E, E->B, CE->A}.... Question 4 Not yet answered Marked out of 2.00 P Flag question Let R(A,B,C,D,E) be a relation with FDs F = {AB-C, CD-E, E-B, CE-A} Consider an instance of this relation that only contains the tuple (1, 1, 2, 2, 3). Which of the following tuples can be inserted into this relation without violating the FD's? (2 points) Select one: 0 (0, 1,...
5. [5 points] Let relation R (A, B, C, D, E) satisfy the following functional dependencies: AB → C BC → D CD → E DE → A AE → B Which one of the following FDs is also guaranteed to be satisfied by R? A. B. BCD → A A-B D. CE → B
Given the following relation schemas and the sets of FD's: a- R(A,B,C,D) F={ABẠC,C7D, D´A, BC+C} b- R(A,B,C,D) F={BẠC, BD, AD>B} C- R(A,B,C,D) F={AB-C, DC+D, CD+A, AD+B} d- R(A,B,C,D) F={AB=C, C+D, D™B, DE} e- R(A, B, C, D, E) F= {AB+C, DB+E, AE>B, CD+A, ECD} In each case, (i) Give all candidate keys (ii) Indicate the BCNF violation Give the minimal cover and decompose R into a collection of relations that are BCNF. Is it lossless? Does it preserve the dependencies?...
Let R(A,B,C,D) be a relation with FDs F = {A—B, AC, C-A, B,C, ABC-D} Which of the following statements is correct ? (2 points) Select one: G = {A-B, B-C, C-A, AC=D } is a canonical cover of F H = { AC, CA, BC,BD} is a canonical cover of F. o F is a canonical cover of itself. O G and H are canonical covers of F. None of the above.
Consider the following relation: R(A,B,C,D,E) The following set of functional dependencies are ture on the relation R: FD: AB -> E, E -> D, AD -> C Which of the following sets of attributes does not functionally determine C? AC ABE BD AE AB
Consider relation R = (A,B,C,D,E) for the following: 2. Is R in 2NF with the following functional dependencies? If not, normalize it. [5 points] A→BC -AD → E в с 3. Are the relations from the answer of question 2 in 3NF? If not normalize it. 5 points]