ACorrect Answer

You would rewrite W with a better key (c,b), since c->a means that c is a better key than a. After that the normalisation is straightforward, and the result is that shown here.

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BIncorrect Answer

This set of relations solves nothing, and does not change the fact that W is still not in BCNF.

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CIncorrect Answer

This set of relations are actually valid when considered seperately, but the resulting query is suboptimal and could result in spurious values (if 'a' contained duplicate values).

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DIncorrect Answer

You cannot delete a directly from the composite key of W, as this would be an illegal operation. You must have 'a' in the key, either directly or indirectly. Note that as c gives a, you could substitute the attribute c every time you use a in a key.

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EIncorrect Answer

You would rewrite W with a better key (c,b), since c->a means that c is a better key than a. After that the normalisation is straightforward.

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A B C D E TELL ME NEXT INDEX
 
 

If the relation W was normalised to BCNF, select the best resulting relationsof that process.

 W(a,b,c,d,e)
 c -> a
  1.  W(c,b,d,e)
     W1(c,a)
    
  2.  W(a,b,c,d,e)
     W1(c,a)
    
  3.  W(a,b,d,e)
     W1(c,a)
    
  4.  W(b,c,d,e)
     W1(c,a)
    
  5. None of the above