Before n is added to the list, the median is 5.5 (the average of 5 and 6). When n is added to the list, the number of elements becomes odd, so the median will be a value directly from the list, not the average of two values. All of the values in the old list are integers and n is an integer, so the new median must be an integer; therefore, 5.5 cannot be the median of the new list.

Considering some possible values of n, we see that in cases where n is less than or equal to 5, the fourth element in the new list would be 5, making the new median 5. In cases where n is greater than or equal to 6, the fourth element in the new list would be 6, making the new median 6. The possible values for the median of the new list are therefore 5 and 6, but not 5.5.

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:

Dtotal = 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.

42.5 is the mean of the data. 13 is the minimum. 83 is the maximum. 70 is the range.

To find the median, list all numbers in order:

13, 21, 25, 37, 51, 52, 58, 83

and find the middle value. In cases like this where there are two middle numbers (37 and 51), find the mean of these two numbers.

(37+51)/2 = 88/2 = 44

To find the median, one must arrange all the heights from the lowest to the highest value and then pick the middle value.

All values: 64 69 60 70 74 72 55 52 58

In order from lowest: 52 55 58 60 64 69 70 72 74

Median: 64

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 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.

You must first find what x is in order to find the median.

To find x, set up the following equation:

To solve the equation, first multiply both sides by 5:

5 + 10 + 12 + 15 + x = 55

Then, add up 5 + 10 + 12 + 15 to get 42:

42 + x = 55

x = 13

Now that you know what x is, you are ready to find the median. To find the median, order the numbers from lowest to highest. The median is the number in the middle.

5, 10, 12, 13, 15

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.

Median is the middle number when the data are ordered from least to greatest. In this case, there are 12 numbers, so, the median is the average of the middle two numbers, which are the 6th and 7th numbers in an ascending order. For the averaged high-temperature, the median is (56+60)/2=58, and for the averaged low-temperature, the median is (35+39)/2=37

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!)