Let's represent the eleven integers as A, B, C, D, E, F, G, H, I, J, and K. Let's assume that if we lined them up from least to greatest, the set would be as follows:

A, B, C, D, E, F, G, H, I, J, K

The median in this set is the middle number. In this case, the middle number will always be F, because there are five numbers before it and five numbers after it.

We are told that the median is 39. Thus F = 39, and our set looks like this:

A, B, C, D, E, 39, G, H, I, J, K

Since we know that all the numbers are consecutive odd integers, the set must consist of the five odd numbers before 39 and the five odd numbers after 39. In other words, the set is this:

29, 31, 33, 35, 37, 39, 41, 43, 45, 47, 49

The question asks for the range of the set, which is the difference between the largest and smallest numbers. In this case, the range is 49 – 29 = 20.

The answer is 20.

First, arrange the numbers from highest to lowest, then subtract the highest number from the lowest number.

The range of a data set is the distance between the smallest number and the largest number. To find the range, you subtract the smallest number (5) from the largest number (13). Therefore, the range is 8.

To find the range of a data set, subtract the smallest number in the set from the largest:

There are a total of 7 + 5 + 3 + 6 = 21 marbles. The probability of picking a yellow marble is 7/21 = 1/3. Then we put it back and choose a blue marble with probability 3/21 = 1/7. We do NOT put this blue marble back, but then we grab for a white. The probability of picking a white is now 6/20 = 3/10, because now we are choosing from 20 marbles instead of 21. So putting it together, the probability of choosing a yellow marble, replacing it and then choosing a blue and a white, is 1/3 * 1/7 * 3/10 = 1/70.

Add up the total number of jellybeans, 20 + 45 + 30 + 5 = 100. 

Divide the number of watermelon jellybeans by the total: 20/100 and reduce the fraction to 1/5.

There are four different types of cake. In this type of problem we want to guarantee we have one of each, so we need to assumbe we have very bad luck. We start with the red velvet since that is the type with the most cakes. If we open those 6 we are not guaranteed to have different ones. Then say we opened all five chocolate cakes, then all four carrot cakes. We still have only three types of cakes but opened 15 boxes. When we open the next box (16) we will be guaranteed to have one of each.

The probability of the point being inside the circle is the ratio of the area of the circle to the area of the square. If we suppose that the circle has radius r, then the square must have side 2r. The area of the circle is πr² and the area of the square is 〖(2r)〗²=〖4r〗², so the proportion of the areas is (πr²)/〖4r〗²  =π/4.

The bowl has 54 marbles, half green and half blue. This gives us 27 green and 27 blue marbles:

27 G / 27 B

John then takes 3 green and 6 blue from the bowl. This leaves the bowl with:

24 G / 21 B

If there are going to be more blue than green marbles after John's 13 marbles, he has to take at least 4 more green marbles than blue marbles, because right now there are 3 less blue marbles. Therefore, we need to take at least 9 green marbles, which would mean 4 or less of the marbles would be blue (8 green and 5 blue would leave us with equal green and equal blue marbles, so it would have to be more than 8 green marbles, which gives us 9 green marbles).

We can also solve this as an inequality. You take the difference in marbles, which is 3, which means you need the difference in green and blue marbles to be greater than 3, or at least 4. You have b + g = 13 and g - b > 3, where b and g are positive integers.

b + g = 13 (Subtract g on both sides of the equation)

b = 13 - g

g - b > 3 (Substitute above equation)

g - (13 - g) > 3 (Distribute negative sign in parentheses)

g - 13 + g > 3 (Add both g variables)

2g - 13 > 3 (Add 13 to both sides of the inequality)

2g > 16 (Divide both sides of the inequality by 2)

g > 8 so g has to be 9 or greater.

If x is chosen at random from the set (4, 6, 7, 9, 11) and y is chosen at random from the set (12, 13, 15, 17) then what is the probability that xy is odd?

Here we have 5 possible choices for x and 4 possible choices for y, giving us 5 * 4 = 20 possible outcomes.

We know that odd times odd = odd; even times even = even; and even times odd = even. Thus we need all of the outcomes where x and y are odd. We have 3 possibilities of odd numbers for x, and 3 possibilities of odd numbers for y, so we will have 9 outcomes of our total 20 outcomes where xy is odd, giving us a probability of 9/20.