The mode is the most occuring value in a set.  $\displaystyle 7$ is the mode because it occurs 6 different times in the set.

The standard deviation tells us how much variation, or dispersion, there is on average between the numbers in a set and their mean.  Therefore Quantity B is greater, because Set B has a larger standard deviation than Set A.

The range is the difference between the largest and smallest numbers of a set. The biggest number is 15 and the smallest is 1, so our range is 15 – 1 = 14.

The median is the middle value in a set of numbers, and the mode is the number that occurs.

In Column A, the median is 6 and the mode is 6, so the sum of the median and the mode of Set G is 12.

The range is the difference between the highest and lowest numbers of the set.

In Column B, the range is 12 – 2 = 10.

Column A is greater than Column B.

The first step of this problem should be to reorder Set Q into numerical order:

$\displaystyle SetQ=[2,5,3,7,8,7]\rightarrow[2,3,5,7,7,8]$

The range of a set is the largest value minus the lowest value, so for this set, the range is $\displaystyle 8-2=6$

The median of a set depends on whether or not there is an even or odd amount of numbers in the set. For a set $\displaystyle [x_1,x_2,...,x_n]$ with $\displaystyle n$ values:

Odd:

$\displaystyle Median=x_{\frac{n+1}{2}}$

Even:

$\displaystyle Median=\frac{x_{\frac{n}{2}}+x_{\frac{n+2}{2}}}{2}$

Since there is an even number of numbers in this set, the median is

$\displaystyle \frac{5+7}{2}=6$

The two quantities are equal.

Given xy = 4y/x and x and y not 0. 

Therefore you are able to divide both sides by 'y' such that:

x = 4/x

Multiply both sides by x:

x² = 4 or x = +2 or –2.

Because of the fact that x could equal –2, the relationship cannot be determined from the information given.

Since $\displaystyle \dpi{100} \small x$ is between 10 and 20, it can be any real number between 100 and 400. Therefore, the relationship cannot be determined since $\displaystyle \dpi{100} \small x$ could fall anywhere between these two limits, including between 399 and 400.

In order to determine the relationship between Quantity A and Quantity B, let's convert both to square roots. In order to do this, we must square Quantity A so it becomes √81 which is equivalent to 9. Now to Quantity B, we must simplify by adding the two values together (25 + 55) to get √80.  

√81 is greater than the √80 because 81 is greater than 80. Thus Quantity A is greater.

$\displaystyle \sqrt[2]{24,300}$

When simplifying the square root of a number that may not have a whole number root, it's helpful to approach the problem by finding common factors of the number inside the radicand. In this case, the number is 24,300.

What are the factors of 24,300?

24,300 can be factored into:

$\displaystyle 24,300: 243\cdot 100$

When there are factors that appear twice, they may be pulled out of the radicand. For instance, 100 is a multiple of 24,300. When 100 is further factored, it is $\displaystyle 10^{2}$ (or 10x10). However, 100 wouldn't be pulled out of the radicand, but the square root of 100 because the square root of 24,300 is being taken. The 100 is part of the24,300. This means that the problem would be rewritten as: $\displaystyle 10\sqrt[2]{243}$ 

But 243 can also be factored: $\displaystyle 24,300: (3\cdot 81)\cdot100$
$\displaystyle 24,300: {\color{Orange} 9}\cdot{\color{Orange} 9}\cdot {\color{Cyan} 10}\cdot {\color{cyan} 10}\cdot 3$
Following the same principle as for the 100, the problem would become
$\displaystyle 9\cdot 10\sqrt[2]{3}$ because there is only one factor of 3 left in the radicand. If there were another, the radicand would be lost and it would be 9*10*3. 
9 and 10 may be multiplied together, yielding the final simplified answer of 
$\displaystyle 90\sqrt[2]{3}$

To solve the equation $\displaystyle \sqrt{180} + \sqrt{125} = ?$, we can first factor the numbers under the square roots.

$\displaystyle \sqrt{2\cdot 2\cdot 3\cdot 3\cdot 5} + \sqrt{5\cdot 5\cdot 5} = ?$

When a factor appears twice, we can take it out of the square root.

$\displaystyle 2\cdot 3\sqrt{5} + 5\sqrt{5} = ?$

$\displaystyle 6\sqrt{5} + 5\sqrt{5} = ?$

Now the numbers can be added directly because the expressions under the square roots match.

$\displaystyle 6\sqrt{5} + 5\sqrt{5} = 11\sqrt{5}$