THIS IS FOR chrissy7905: Consider a set S = {2, 4, 6, 8, x, y} with distinct elements. If x and y are both prime numbers and 0 < x < 40 and 0 < y < 40, which of the following MUST be true? A. I only
B. I and II only
C. I and III only
D. III only
E. I, II, and III

the answer would be e

Step-by-step explanation:S = {2, 4, 6, 8, x, y}

x and y are both prime numbers and 0 < x < 40 and 0 < y < 40

I. The maximum possible range of the set is greater than 33.

The biggest prime number < 40 is 37 . To obtain the max possible range either x or y has to be 37

Range = 37-2 = 35

True

II. The median can never be an even number.

S = {2, 4, 6, 8, x, y}

If x and y are prime numbers greater than 8 , then median will be (6+8)/2 = 7

If x and y are 3 and 5 , then set S = {2,3,4,5,6,8} , then median will be = 9/2

If x = 3 , then set S = { 2,3,4,6,8, y} , then median will be = 5

If x = 5 , then set S = { 2,4,5,6,8,y} , then median will be = 11/2

If x = 7 , then set S ={ 2,4,6,7,8,y} , then median will be = 13/2

Median will either be an odd number or a decimal .

True

III. If y = 37, the average of the set will be greater than the median.

S = {2, 4, 6, 8, x, 37}

The average will be least when x is smallest

S = {2, 3, 4, 6, 8, 37}

If x=3 , then average will be 60/6 = 10 and median will be 5 .

The median will be highest when x>8 , median = 7 , now the average will be more than 10 .

im not good at theseee ;( give me any algebra equation and i can do it in like 5 seconds heh

Step-by-step explanation:

p = 6s

step-by-step explanation: