This question is asking to find the mean of the speed travelled by the Johnson family. To do this you need to add the speeds and divide them by the number of legs in the trip.

So:

\(\displaystyle \frac{45+70+45}{3}=\frac{160}{3}=53.33\)

Add all the numbers and divide by the number of numbers in the dataset.  There are four numbers in this dataset.

\(\displaystyle \frac{3+(-9)+6+(-8)}{4}\)

Simplify the numerator.

\(\displaystyle \frac{-6+6-8}{4} = \frac{-8}{4} =-2\)

The answer is \(\displaystyle -2\).

Set up the equation that will show the mean of Billy's four exams.  The left side of the equation must equal to 80 for his final average.

\(\displaystyle \frac{65+60+x+x}{4} = 80\)

Solve for \(\displaystyle x\).  This is Billy's bare minimum to achieve his desired score.

\(\displaystyle 65+60+x+x = 320\)

\(\displaystyle 2x +125 = 320\)

Subtract 125 on both sides.

\(\displaystyle 2x =195\)

Divide by 2 on both sides.

\(\displaystyle \frac{2x}{2} =\frac{195}{2} = 97.5\)

Billy will need a minimum of \(\displaystyle 97.5\) on both exams.

Recall that finding the average requires you to add up the terms and then divide by the number of terms. Since we are trying to find the score of the sixth test, we have to remember that there are 6 terms. Therefore, our equation looks like this: \(\displaystyle \frac{98+89+92+75+90+x}{6}=87\). Then, solve for x, which is 78.

Find the mean of the following data set:

\(\displaystyle 1,43,117,42,2952,54,18,97,112,23,117\)

To find the mean, we simply need to sum our terms and divide by the number of terms. Our total number of terms is 11:

\(\displaystyle 1+43+117+42+2952+54+18+97+112+23+117=3576\)

\(\displaystyle \frac{3576}{11}=325.\bar{09}\approx325\)

So our mean is 325

Find the mean of this data set:

\(\displaystyle 145,57,223,76,453,123,979,57,76,233,435,76\)

To begin, let's put our numbers in increasing order:

\(\displaystyle 57,57,76,76,76,123,145,223,233,435,453,979\)

Next, find the sum of our terms, and divide that by the number of terms.

\(\displaystyle \frac{57+57+76+76+76+123+145+223+233+435+453+979}{12}\approx244\)

So our answer is 244

Find the mean of the following data set:

\(\displaystyle 66,123,44,78,99,67,143,44,107,12,578,12,67,367,44\)

Let's begin by putting our data in increasing order:

\(\displaystyle 12,12,44,44,44,66,67,67,78,99,107,123,143,367,578\)

The mean of our data set will be the same as the average. Find the sum of the terms, then divide by the number of terms (15 in this case)

\(\displaystyle \\12+12+44+44+44+66+67+67+78+99+107+123+143+367+578=1851\)

\(\displaystyle \mu=\frac{1851}{15}=123.4\approx123\)

So our answer is 123.

The median must be 15. 15 is the middle number weather the 5th number if less than 12 or more than 15. If the median is 15 the mean must also be 15.  

Once we know that we use the equation of the mean to solve the problem.

\(\displaystyle \frac{15+17+12+15+x}{5} = 15\)

 Now solve for x

\(\displaystyle \frac{59+x}{5} = 15\)

Multiply by 5 on both sides

\(\displaystyle 59+x = 75\)

Subtract 59 on both sides 

\(\displaystyle x= 16\)

To find the mean of these numbers you must add them all up and divide by the total number of players.

You get:

\(\displaystyle \frac{103}{10}=10.3\)   

which rounds to \(\displaystyle 10\).

The mean of student A is \(\displaystyle \frac{65+60+76+44+90}{5} = 67\)

The range of studnet A is \(\displaystyle 90-44 = 46\)

The mean of student B is \(\displaystyle \frac{70+63+74+60+102}{5} = 73.8\)

The range of studnet B is \(\displaystyle 102-60 = 42\)

The range of student A is more the the range of student B.