The number of ways to select three elements from a set of eight is the number of combinations of three elements chosen from eight:

$\displaystyle C(8,3) = \frac{8!}{3! (8-3)!}$

$\displaystyle = \frac{8!}{3!5!}$

$\displaystyle = \frac{8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 }{3 \times 2 \times 1 \times 5 \times 4 \times 3 \times 2 \times 1 }$

$\displaystyle = \frac{40,320 }{720}$

$\displaystyle = 56$

20 can be factored as:

I) $\displaystyle 1 \times 20$

II) $\displaystyle 2 \times 10$

III) $\displaystyle 4 \times 5$

The positive difference of the factors can be any of:

$\displaystyle 20 - 1 = 19$

$\displaystyle 10 - 2 = 8$

$\displaystyle 5 - 4 = 1$

Of the four choices, only 8 is possible.

$\displaystyle \pi$ is a well-known irrational number and is not the correct choice. 

$\displaystyle \sqrt{\frac{8}{4}} = \sqrt{2}$. The square root of an integer is rational only if it is itself an integer; calculation yields a result of 1.414... This is not the correct choice.

$\displaystyle \sqrt{\frac{16}{4}} = \sqrt{4} = 2$, since $\displaystyle 2 ^{2} = 2 \times 2 = 4$. This is an integer and is not the correct choice.

$\displaystyle \left ( \frac{3}{2} \right ) ^{2} = \frac{3}{2} \cdot \frac{3}{2} = \frac{9}{4}$, so, by definition, $\displaystyle \sqrt{\frac{9}{4}} = \frac{3}{2}$. This is rational and is therefore the correct choice.

$\displaystyle - \frac{5}{6} = -(5 \div 6) = - 0.8333...$, which is not an integer. It is not a whole number either, as the whole numbers consist of all (nonnegative) integers.

$\displaystyle - \frac{5}{6} = - 5 \div 6$. As the quotient of integers, it is rational. 

Therefore, $\displaystyle - \frac{5}{6}$ belongs to the set of rational numbers but not to the other two sets. The correct response is one.

A number in base sixteen has powers of sixteen as its place values; starting at the right, they are $\displaystyle 1, 16 ^{1} = 16, 16 ^{2} = 256, 16 ^{3} = 4,096$.

The lowest base sixteen number with three digits would be

$\displaystyle 100 _{\textrm{sixteen}} = 1 \times 16^{2} = 256$ in base ten.

The lowest base sixteen number with four digits would be

$\displaystyle 10000 _{\textrm{sixteen}} = 1 \times 16^{4} = 4,096$ in base ten.

Therefore, a number that is expressed as a three-digit number in base sixteen must fall in the range

$\displaystyle 256 \leq N \leq 4,095$.

Of the four numbers listed, 1,000 falls in that range.

The set of whole numbers comprises 0 and the so-called natural, or counting, numbers; that is, it is the set $\displaystyle \left \{ 0, 1, 2, 3, 4,...\right \}$. $\displaystyle -7$ is not one of these numbers.

The set of integers comprises these numbers as well as their (negative) opposites; that is, it is the set $\displaystyle \left \{ ...,-4, -3, -2, -3, 0, 1, 2, 3, 4,...\right \}$. $\displaystyle -7$ is one of these numbers.

A prime number is a positive integer which has exactly two factors - 1 and the number itself. 36, 38, and 39 can each be shown to have at least one other factor:

$\displaystyle 36 \div 2 = 18$, so 2 and 18 are factors of 36, making 36 not prime.

$\displaystyle 38 \div 2 = 19$, so 2 and 19 are factors of 38, making 38 not prime.

$\displaystyle 39 \div 3 = 13$, so 3 and 13 are factors of 39, making 39 not prime.

37, however, cannot be evenly divided by any number except for 1 and itself, as can be seen below:

$\displaystyle 37 \div 2 = 18 \textrm{ R }1$

$\displaystyle 37 \div 3 = 12 \textrm{ R }1$

$\displaystyle 37 \div 4 = 9 \textrm{ R }1$

$\displaystyle 37 \div 5 = 7 \textrm{ R }2$

$\displaystyle 37 \div 6 = 6 \textrm{ R }1$

We do not need to go higher, since any higher possible factors have square greater than 37. 37 has been proved a prime number.

An integer is an element of the set of numbers 

$\displaystyle \left \{ ...-3, -2, -1, 0, 1, 2, 3 ,... \right \};$

that is, the so-called natural, or "counting", numbers, their (negative) opposites, and 0. $\displaystyle -4$, 2, $\displaystyle -8$, and 1 are elements of this set; $\displaystyle \frac{2}{3}$ and $\displaystyle \sqrt{2}$ are not. The correct response is 4.

Remember that a complex number is any number that includes $\displaystyle i$ or $\displaystyle \sqrt{-1}$.

$\displaystyle \frac{19}{91} \approx 0.208791$

$\displaystyle \sqrt{19}$ is irrational but real.  

Similarly, $\displaystyle \sqrt{91 - 19} = \sqrt{72} = \sqrt{9 * 8} = \sqrt{3^{2} * 2^{}3} = 3 * 2 *\sqrt{2} = 6\sqrt{2}$ which is irrational but real.

A prime number is a number that is evenly divisible by $\displaystyle 1$, the number itself, and no other number. 

$\displaystyle $6\hspace{1mm}\div\hspace{1mm}2=3$

$\displaystyle $8\hspace{1mm}\div\hspace{1mm}2=4$

$\displaystyle $9\hspace{1mm}\div\hspace{1mm}3=3$

$\displaystyle 7$ is the only number that is not evenly divisible by a number other than $\displaystyle 1$ and the number itself.