The probability of picking a spearmint or a cinnamon is the addition of probability of picking a spearmint and the probability of picking a cinnamon

\(\displaystyle \frac{3}{14} + \frac{6}{14} = \frac{9}{14}\) not \(\displaystyle \frac{1}{2}\)

Step 1: We need to find out how many outcomes there will be.
If we roll a coin three times, there are \(\displaystyle 2^3\) outcomes.
If we roll a coin \(\displaystyle n\) times, there will be \(\displaystyle 2^n\) outcomes.

Step 2: Find all the outcomes.

The outcomes here are: HHH, HHT, HTH, HTT, THH, THT, TTH, and TTT.

Step 3: In the list of outcomes, count how many times the letter H appears at least once.

The letter "H" appears in HHH, HHT, HTH, HTT, THH, THT, and TTH.

The letter "H" appears in \(\displaystyle 7\) of the \(\displaystyle 8\) outcomes.

The probability of getting at least one H in the outcomes in Step 2 is \(\displaystyle \frac{7}{8}\).

Step 1: Let's answer a smaller problem. How many ways can I toss one coin?

There are two ways, either I get Heads or Tails.

Step 2: How about two coins?

There are four ways... They are, HH, HT, TH, and TT

Step 3: How many different combinations for three coins?
Let's List them:
HHH, HHT, HTH, THH, TTT, THT, TTH, HTT

There are \(\displaystyle \large 8\) different combinations.

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.

\(\displaystyle x=17\)