There exists only one possible 3NF decomposition for any relation schema and any set of functional dependencies.Question 4Select one:TrueFalse

Consider relation StudentInfo(sID, name, dorm, major) with functional dependency sID → name and multivalued dependency sID ↠ dorm. What schema would be produced by the 4NF decomposition algorithm?

Consider the universal relation R = {A, B, C, D, E, F, G, H, I, J} and the set of functional dependencies F = { {A, B}→{C}, {A}→{D, E}, {B}→{F}, {F}→{G,H}, {D}→{I, J} }. What is the key for R? Decompose R into 2NF and then 3NF relation

What is TRUE about the Third Normal Form (3NF)?a.All of the answerb.Data duplication is reduced by using 3NFc.When a relation is in 2NF and does not contain transitive partial dependencies, it will be in 3NFd.It helps maintain the integrity of the data

Let R(A,B,C,D,E,P,G) be a relational schema in which the following FDs are known to hold:AB->CDDE->PC->EP->CB->GThe relation schema R isin 3NF, but not in BCNFin BCNFin 2NF, but not in 3NFnot in 2NF

Consider a relation schema R = {A, B} and a set of functional dependencies Σ = {A → B}, AB must be a key of R.Question 5Select one:TrueFalse