Statement I must be true because if M had an even number of consecutive integers, then its average (and median) would be the average of two consecutive integers, which is a decimal value rather than an integer, and therefore cannot be in set M, which contains only integers.

To check Statement II, consider some simple possible sets for M such as {0,1,2}. We see that in a set with an odd number of items, like M, the median is always the middle element. We also see that in a set with an odd number of consecutive integers (such as {0,1,2}), the average of the set will always be the exact middle element, too. Therefore, the average and the median must be equal, and Statement II must be true.

Finally, we can find a counterexample to statement III to show that it does not have to be true. If set M is  {0,1,2}, we see that there is no unique mode (in fact, the only time set M could have a unique mode is when it only has one element!)

To find the median, you need to find the middle number of the ordered list.  Since there are four numbers in the list, you need to average the two middle numbers.

Ordered list = {x – 8, x – 4, x + 2, x + 6}

((x – 4) + (x + 2))/2 = (2x – 2)/2 = x – 1

The easiest way to solve this is to find the total number of miles traveled by the car and divide that by the total time travelled.

Recall that D = rt; therefore, for each of these three periods, we can calculate the distance and sum those products:

D_total = 60 * 3 + 55 * 2 + 40 * 3 = 180 + 110 + 120 = 410

The total amount of time travelled is: 3 + 2 + 3 = 8

Therefore, the average rate is 410 / 8 = 51.25 miles per hour.

Let's let a and b stand for the smallest and largest values in S, respectively. Because T is equal to the square of the elements in S, then the smallest and largest values in T will be equal to a² and b² , respectively.

We are told that the range of T is equal to 560. The range of a set of numbers is the difference between the largest and smallest, which in this case we can represent as b² – a². We can now write the following equation:

b² – a² = 560

Noice that b² – a² can be factored as (b – a)(b + a), which is the formula for the difference of squares. 

(b – a)(b + a) = 560

Notice that the value b – a represents the range of S, because it is the difference between the largest and smallest values of S.

We are told that S consists of six consecutive odd integers. We can use this information to find the range of S. Because they are consecutive odd integers, each member in S will be two larger than the one before it. Thus, we could represent S as follows:

a, a + 2, a + 4, a + 6, a + 8, a + 10

The range of S is equal (a + 10) – a = 10. Thus, b – a must equal 10.

Let's go back to the equation (b – a)(b + a) = 560. We can replace b – a with 10 and solve for b + a.

(10)(b + a) = 560

Divide both sides by 10.

b + a = 56

So, we know that b – a = 10, and b + a = 56. We can solve this system of equations to find the values of a and b, which will give us the smallest and largest values in S.

Let's solve the first equation in terms of b.

b – a = 10

Add a to both sides.

b = 10 + a

Now we can substitute 10 + a in for b in the second equation.

(10 + a) + a = 56

10 + 2a = 56

Subtract 10 from both sides.

2a = 46

Divide both sides by 2.

a = 23

b = 10 + a = 10 + 23 = 33

The smallest value of S is 23, and the largest value is 33. Thus, S consists of the following set:

23, 25, 27, 29, 31, 33

We are told that T is equal to the squares of the elements of S. Thus, T consists of the following numbers:

529, 625, 729, 841, 961, 1089

We are asked ultimately to find the median of T. The median of any set is the middle number. In this case, T has six elements, so there are actually two numbers in the middle: 729 and 841. To find the median, we will take the average of the two middle numbers. The average of two numbers is equal to their sum divided by two.

median of T = (729 + 841)/2 = 1570/2 = 785.

The answer is 785. 

The set consists of eight consecutive odd integers. Let's call the integers, in order from least to greatest, A, B, C, D, E, F, G, and H. We are told that the median of the set is 22. The median in a set of numbers is the number in the middle. In the set of A, B, C, D, E, F, G, and H, the numbers represented by D and E would be in the middle. When we have two members in the middle, the median is equal to their average. Thus, 22 is equal to the average of D and E. In other words, 22 is the average of the 4th and 5th numbers in the set. Since 22 is halfway between 21 and 23, it makes sense that the 4th and 5th numbers would have to be 21 and 23, respectively. Thus, our set looks like this:

A, B, C, 21, 23, F, G, H

Beacuse the set consists of consecutive odd integers, each integer is two larger than the one before it. Therefore, this is the set of integers:

15, 17, 19, 21, 23, 25, 27, 29

The problem asks us to find the largest integer in the set, which is 29.

The answer is 29.

To find the median, arrange the numbers from lowest to highest then find the middle number.

There are two numbers in the middle in this set ( and ).

In this case, the median is 6 but you would typically find the average of the two numbers in the middle.

The median is the "middle number" in a sorted list of number; therefore, reorder the numbers in numerical order: 7, 8, 12, 15, 19, 20. Since x is the middle number, it must come between 12 and 15, leaving 14 as the only viable choice.

The median is the coin which has an equal number of coins of lesser or equal value and greater or equal value on either side of it.  The median therefore falls to one of the nickels (62 quarters, dimes and nickels above it and 62 nickels and pennies below it).  The value of a nickel is 5 cents.

We first remove any 94%+ scores.  For tests scored out of 50, a 94% is a 47.  Thus anything 47 or greater will be removed.  We see that this leaves us with:

For the second half, we have tests scored out of 40.  94% of 40 gives us 37.6, thus we will only remove tests that were a 38 or better.  This leaves us with:

Now we have to convert them to percentiles.  We end up with the lists as follows. The raw scores are in parentheses:

We then order it by percentile (removing raw scores, as they are unnecessary):

We notice that the median must be between the 78 and the 87.5 percentile scores. 31 out of 40 gives us 77.5%, which is below the score associated with an earlier test.  Thus this cannot be the answer.  The remaining 4 scores are all higher than 78%, and the largest of them is equal to 87.5 (the highest bound). 

The median is the middle value in an ordered set of numbers. In this instance, we have 8 numbers, so the median is the average of the 4th and 5th number:

The median for this set is 3.5. Now let's find the mean, or the average, of the set:

Now that we have the mean and median, we can finish the question. The question asks for the mean (3.375) minus the median (3.5):