Since the mean is the average of the data, we must calculate the average of Brandon's runs. In order to do this, we first add up all the data points.

\(\displaystyle 12 mi+14 mi+13 mi+12 mi+14 mi+16 mi+15 mi+9 mi+14 mi+12 mi+10 mi+11 mi+14 mi+12 mi= 178 mi\)

Next we divide by the total number of days that Brandon runs, \(\displaystyle \small 14\).

\(\displaystyle \small \frac{178}{14}=12.7142857\)

We can round this to the nearest tenth, \(\displaystyle \small 12.7\: miles\). 

Add up the 8 test scores and divide by 8 to figure out the mean (83.5). Now figure out the average between the highest and lowest scores

\(\displaystyle \frac{96+71}{2}=83.5\) 

Since the dropped tests have the same average as the original 8 tests, Allie's score would not change. The correct answer is that the two columns are equal.

Note: If you are good at mental math, here is a quick way to figure out the average of the 8 tests:

First, estimate the average (say, 80). Next, go through the numbers and add or subtract the difference between the number and 80.

79 is 1 below 80.

77 is 3 below 80 (so now we're 4 below 80).

As we continue through the numbers, we get

12 above, 3 above, 8 above, 14 above, 24 above, 28 above.

So the 8 scores are cumulatively 28 above an average score of 80. Now, simply divide 28 by 8 and add this to 80.

\(\displaystyle 3.5+80=83.5\)

You do not need to take the two means; just add the scores, since each sum will be divided by 5 to get the students' means.

Kent's total: \(\displaystyle 83+ 91+ 88+ 85+ 80 = 427\)

Janice's total: \(\displaystyle 87+ 87+ 84+85+92 = 435\)

Janice's sum is higher; so is her mean score.

To get the mean of the students' weights, divide the sum of the weights by the number of students; here, this is 

\(\displaystyle 1,980 \div 12 = 165 < 175\)

(B) is greater.

For the mean of the eleven data elements to be 42, their sum divided by 11 must be 42. Equivalently, the sum of the elements must be \(\displaystyle 42 \times 11 = 462\). 

The sum of the ten known elements is 

\(\displaystyle 20+ 30+ 30+40+40+ 40+40+50+ 50+ 60 = 400\)

Therefore, the unknown element must be \(\displaystyle 462-400 = 62\)

Add the weights, then divide the sum by six:

\(\displaystyle 145+172+166+159+153+201=996\)

\(\displaystyle 996 \div 6 = 166\)

There are six elements, so to find the mean, add the elements and divide by six:

\(\displaystyle \left ( 2+4+8+16+32+64 \right )\div 6 = 126 \div 6 = 21\)

One ton is equal to 2,000 pounds.

The mean of the weights of the students is the total of their weights divided by the number of students.

Let \(\displaystyle T\) be the total of their weights.

\(\displaystyle 186 = T \div 11\)

\(\displaystyle T = 186 \cdot11 = 2,046 > 2,000\)

(A) is greater.

You can find the average of a set of numbers by adding up the numbers and dividing by how many numbers are in the data set. Since David is trying to acheive an average of 85 by his fourth test, we should multiply 85 by 4.

\(\displaystyle \text{average}=85=\frac{\text{total}}{4}\)

\(\displaystyle \text{total}=85\times4 = 340\)

The result is the number that the four tests should total in order to achieve an average of 85. Then, add up the test scores Daid has recieved so far:

\(\displaystyle 77 + 92 + 80 = 249\)

If you subtract this number (the total from the test scores so far) from 340, the result will be the score needed to achieve the 85 average.

\(\displaystyle 340 - 249 = 91\)

David must score a 91 on his fourth test.

You can find the average of a set of numbers by adding up all the numbers in that set and dividing by how many numbers you added total. Here is a summary of our set:

Monday: 2 miles

Tuesday: 3 miles

Wednesday: 0 miles

Thursday: 3 miles

Friday: 2 miles

Keep in mind that we are looking for the average number of miles June ran Monday through Friday, and that there are 5 days in that period. You must also account for the 0 miles that June ran on Wednesday. 

\(\displaystyle \text{average}=\frac{\text{total miles}}{\text{total days}}=\frac{2+3+0+3+2}{5}=\frac{10}{5}\)

June ran a total of 10 miles during that 5 day period. You must then divide by the number of days in order to get the average.

\(\displaystyle 10 \div 5 = 2\)