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April 16, 2014

April 16, 2014

Posted by **mary** on Sunday, February 20, 2011 at 2:58pm.

A = {r, i, s, k, e, d}, B = {r, i, s, e}, C = {s, i, r}

(Points : 2)

B A and B C

A C and C B

B A and C A

C A and B C

A B and C B

- math -
**MathMate**, Sunday, February 20, 2011 at 3:25pmFor posts with mathematical symbols, please post with encoding Western ISO-8859-1 (Firefox) or Western-Europe(ISO) (Internet Explorer).

The above post in Western ISO-8859-1 is as follows:

Given the following sets, select the statement below that is true.

A = {r, i, s, k, e, d}, B = {r, i, s, e}, C = {s, i, r}

(Points : 2)

B ⊂ A and B ⊂ C

A ⊂ C and C ⊂ B

B ⊆ A and C ⊂ A

C ⊆ A and B ⊆ C

A ⊆ B and C ⊆ B

B ⊂ A means that every element found in B is found in A, but B≠A, i.e. B is a proper subset of A.

If B*can*equal A, the symbol ⊆ is used.

For example, elements in B are r,i,s and e. Each of the four elements are also found in A which has a higher cardinality than B. So B ⊂ A is true.

On the other hand, the cardinality of B is 4, while that of C is 3. So B⊂C is not possible. So the first statement is false.

You can continue this way, and post your results for checking if you wish.

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