Statement (1) allows us to find B:

B = S + 3 = {4 , 6 , 10} Therefore:

S + B = {1+4 , 1+6 , 1+10 , 3+4 , 3+6 , 3+10 , 7+4 , 7+6 , 7+10}

S + B = {5 , 7 , 9 , 11 , 13 , 17}. SO statement (1) is sufficient to find S+B

Statement (2) does not give us enough information to find B. It could be any set between  {4 , 6 , 10} and {1 , 3 , 4 , 6 , 7 , 10} if it includes the same numbers as S. Therefore Statement 2 is not sufficient.

Statement 1 alone only eliminates Regions I and II (since whole numbers are nonnegative integers); negative numbers can be found in Regions III, IV, and V.

Statement 2 alone states that  is an integer that as not a whole number - that is,  is a negative integer. Since , as a consequence, . , the product of a positive integer and a negative integer, is a negative integer, and it would be placed in Region III, which comprises exactly the negative integers.

Region IV comprises the rational numbers that are not integers. A number is rational if and only if it can be expressed as the quotient of integers. 

From Statement 1 alone, it can be inferred that  is rational, and that it is not an integer. Since , it follows that . However, this is not sufficient to narrow it down completely.

For example:

If , then , a natural number, putting it in Region I.

If , then , a rational number but not an integer, putting it in Region IV.

From Statement 2 alone, it can be inferred that  is rational, and that it is not an integer. From , it follows that . The nonzero rational numbers are closed under division, so  must be a rational number. However, since  is not an integer,  cannot be an integer, since the integers are closed under multiplication. Therefore, Statement 2 alone proves that  belongs in Region IV.

Region I comprises the natural numbers - 

From Statement 1 alone,  is a natural number; since , it follows that  is the difference of a natural number and 7 - that is, 

 could be in any of three regions - I, II, or III.

Conversely, from Statement 2 alone,   is the sum of a natural number and 7 - that is,

 must be a natural number and it must be in Region I.

Assume Statement 1 alone. It cannot be determined what region  is in. 

For example, suppose , which is in Region I (the set of natural numbers, or positive integers). It is possible that , putting it in Region I, or , putting it in Region III (the set of integers that are not whole numbers - that is, the set of negative integers).

Assume Statement 2 alone. It cannot be determined what region  is in. 

For example, suppose , which is in Region III; then , which is also in Region III. But suppose ; then , which, as an irrational number, is in Region V.

Now assume both statements. Then  has an integer as a square and an integer as a cube.  must either be an integer or an irrational number. But 

, making it the quotient of integers, which is rational. Therefore,  is an integer. Furthermore, its cube is negative, so  is negative. The two statements together prove that  is a negative integer, which belongs in Region III.

A set with  elements has exactly  subsets in all, and  proper subsets (every subset except one - the set itself). 

From Statement 1, since  has  subsets, it follows that it has 6 elements. From Statement 2, since  has 63 proper subsets, it has 64 subsets total, and, again, 6 elements. Either statement alone is sufficient.

Statement 1 alone gives insufficient information. For example, if , then:

 or 

Since , it is unclear which of  and  is greater, if either.

Statement 2 gives insufficient information; if  is positive,  is negative, and vice versa.

Assume both to be true. The two statements form a system of equations that can be solved using substitution:

Case 1: 

Case 2: 

This equation has no solution.

Therefore, the only possible solution is . Therefore, it can be concluded that .

To answer this question, we must understand the vocabulary.  The range of a set is the difference between the largest and smallest numbers.  The mode of a set is the number which comes up most frequently.  Neither the range nor the mode, nor both, will help us solve for the average, however.

standard deviation = 

Percentile = 

standard deviation of the sample mean is:

95%  for 14 d.o.f. =

confidence interval is