The mean can be found in the same way as the average of a group of numbers. To find the average, use the following formula:

\(\displaystyle Mean=\frac{\textup{sum of data points}}{\textup{total number of data points}}\)

So, if our set consists of 

\(\displaystyle S=(13,14,177,65,87,23,45,78,23)\)

We will get our mean via:

\(\displaystyle Mean=\frac{13+14+177+65+87+23+45+78+23}{9}=58.\bar{3}\)

So our answer is

\(\displaystyle 58.\bar{3}\)

To find the sum of the four numbers, just multiply four and the average. By multiplying the average and number of terms, we get the sum of the four numbers regardless of what those values could be.

\(\displaystyle 35*4=140\) Since Quantity A matches Quantity B, answer should be both are equal. 

Let's look at a case where \(\displaystyle z=70\).

Let's have \(\displaystyle y\) be \(\displaystyle 109\) and \(\displaystyle x\) be \(\displaystyle 1\). The sum of the three numbers have to be \(\displaystyle 3*60\) or \(\displaystyle 180\). 

The average of \(\displaystyle x,y\) is \(\displaystyle \frac{110}{2}\) or \(\displaystyle 55\). The avergae of \(\displaystyle y,z\) is \(\displaystyle \frac{179}{2}\) or \(\displaystyle 89.5\).

This makes Quantity B bigger, HOWEVER, what if we switched the \(\displaystyle y\) and \(\displaystyle x\) values. 

The average of \(\displaystyle x,y\) is still \(\displaystyle \frac{110}{2}\) or \(\displaystyle 55\). The avergae of \(\displaystyle y,z\) is \(\displaystyle \frac{71}{2}\) or \(\displaystyle 35.5\).

This makes Quantity A bigger. Because we have two different scenarios, this makes the answer can't be determined based on the information above.

Let's add each expression from each respective quantity

Quantity A: \(\displaystyle 2+3+7+4+x=16+x\)

Quantity B: \(\displaystyle 1+6+5+8+y=20+y\)

Since \(\displaystyle x>y\) we will let \(\displaystyle x=10\) and \(\displaystyle y=6\). The sum of Quantity A is \(\displaystyle 26\) and the sum of Quantity B is also \(\displaystyle 26\). HOWEVER, if \(\displaystyle y\) was \(\displaystyle 7\), that means the sum mof Quantity B is \(\displaystyle 27\). With the same number of terms in both quantities, the larger sum means greater mean. First scenario, we would have same mean but the next scenario we have Quantity B with a greater mean. The answer is can't be determined from the information above. 

To figure out which Quantity is greater, let's find the highest possible mean in Quantity A. We should pick the \(\displaystyle 5\) biggest numbers which are \(\displaystyle 4, 5, 7, 8, 9\). The mean is \(\displaystyle \frac{4+5+7+8+9}{5}=\frac{33}{5}=6.6\). This is the highest possible mean and since Quantity B is \(\displaystyle 7\) this makes Quantity B is greater the correct answer.

To find the mean, add the terms up and divide by the number of terms.

\(\displaystyle \frac{5+10+18+96}{4}=\frac{129}{4}=33.25\)

To find the mean, add the terms up and divide by the number of terms.

\(\displaystyle \frac{7+15+25+36+x}{5}=20\) Then add the numerator.

\(\displaystyle \frac{83+x}{5}=20\) Cross-multiply.

\(\displaystyle 83+x=100\) Subtract \(\displaystyle 83\) on both sides.

\(\displaystyle x=17\)

If the average of \(\displaystyle x\) and \(\displaystyle y\) is \(\displaystyle 40\), then the sum must be \(\displaystyle 80\). 

\(\displaystyle \left(\frac{x+y}{2}=40; x+y=80\right)\)

If we add the sum of \(\displaystyle x, y, z\) we get \(\displaystyle 80+10\) or \(\displaystyle 90\).

To find the average of the three terms, we divide \(\displaystyle 90\) and \(\displaystyle 3\) to get \(\displaystyle 30\). 

The first ten prime numbers are \(\displaystyle 2, 3, 5, 7, 11, 13, 17, 19, 23, 29\). Prime numbers have two factors: \(\displaystyle 1\) and the number itself. Then to find mean, we add all the numbers and divide by \(\displaystyle 10\).

\(\displaystyle \frac{2+3+5+7+11+13+17+19+23+29}{10}=\frac{129}{10}=12.9\)

There are two methods.

Method 1:

Since the average of \(\displaystyle 7\) consecutive number is \(\displaystyle 8\), we can express this as:

\(\displaystyle \frac{x+x+1+x+2+x+3+x+4+x+5+x+6}{7}=8\)

\(\displaystyle \frac{7x+21}{7}=8\) Cross-multiply

\(\displaystyle 7x+21=56\) Subtract \(\displaystyle 21\) on both sides then divide both sides by \(\displaystyle 7\)

\(\displaystyle x=5\) Since we are looking for the second term, just plug into expression \(\displaystyle x+1\). That means answer is \(\displaystyle 5+1\) or \(\displaystyle 6\). 

Method 2:

Since the average of \(\displaystyle 7\) consecutive number is \(\displaystyle 8\), this means the median is also \(\displaystyle 8\). Consecutive means one after another and with each term having the same difference, we know the mean equals the median. Since there are \(\displaystyle 7\) terms, the median will have \(\displaystyle 3\) terms left and right of it. Then, the series will go \(\displaystyle 5, 6, 7, 8, 9, 10, 11\). The second term is \(\displaystyle 6\).