It is easier to count the tiles that are vowels or blanks than it is to count the tiles marked with consonants. Keeping in mind that one of each vowel has been added, the number of each vowel is as follows:

A: 10 tiles

E: 13 tiles

I: 10 tiles

O: 9 tiles

U: 5 tiles

Blanks: 2 tiles

Add these:  tiles not marked with consonants. 

Since the five unique tiles - "J", "K", "Q", "X", and "Z" - were replaced by five vowel tiles, the number of tiles remained constant at 100. Therefore, there are 

consonant tiles; consonants therefore make up more than one half of the tiles, and the probability that a randomly drawn tile will be a consonant is greater than .

The number of tiles with each vowel is as follows:

A: 9 tiles

E: 12 tiles

I: 9 tiles

O: 8 tiles

U: 4 tiles

This is a total of  vowel tiles in the complete set. Considering the two blanks, there are 

consonant tiles.

There are five unique tiles - "J", "K", "Q", "X", and "Z". Removing them leaves  consonant tiles. Since the removal of seven tiles and their replacement with seven others leaves the number of tiles in all at 100, at the very least, there are 51 consonant tiles in the new set - more than half.

The costs of the beverages will be as follows:

Add these amounts:

20% of this is 

, which rounds to $5.30.

The size of the regions do not matter in this problem; it is the central angles of the sectors that are relevant. The best way to understand this is to look at the spinner again, but after cutting out the central circle:

If we rearrange the sectors, we see the following:

The probability of the spinner landing in red exceeds that of it landing in green.

The median of seven (an odd number) integers is the one in the middle when the numbers are arranged in ascending order; in this case, it is the fourth lowest. Since the seven integers are consecutive, the lowest integer is three less than the median, or .

(a) The mean of this data set is the sum divided by 10:

(b) The median of a data set with ten elements is the arithmetic mean of the fifth-highest and sixth-highest elements:

The set, arranged, is 

The median is 

This makes (a) the greater quantity

Each data set has ten elements, so the median in each case is the arithmetic mean of the fifth-highest and sixth-highest elements. In each data set, these elements are 10 and 10, so the median of each set is 10. Therefore, both quantities are equal.

(a) The mean of this set is .

(b) Since there are  elements, the median of this set is the seventh-highest number, which is .

Therefore (b) is the greater quantity.

The first ten primes form the data set:

(a) Add these primes, and divide by :

(b) The median of a data set with ten elements is the arithmetic mean of the fifth-highest and sixth-highest elements. These are  and , so the median is 

.

(a) is the greater quantity.

(a) The mean of nine elements is , so, if  is their sum,  and . The sum of the new data set is . Since the new set has  elements, its mean is .

(b) The median of the nine elements is , so, when they are ranked, the fifth-highest element is . Since  is less than  and is greater than , when they are added to the set,  is the sixth-highest of eleven elements, which is the median.

Therefore, both are equal to .