In order to answer this question correctly, we need to solve for the mean, median, and mode of this data set. 

To begin, let's sort the data from least to greatest:

Now that our data is ordered from least to greatest, we can solve for the median:

Remember, the median is the middle most number when a data set is ordered from least to greatest.

The median for this data set is 

Next, we can look at our data set to determine the mode:

The mode for this set is 

Remember, the mode is the number in a set that appears most often. 

Finally, we can solve for the mean:

Remember, the mean of a data set is the average of the numbers in a data set. 

The mean for this data set is 

Now that we've done our calculation we should have:

Median: 

Mode: 

Mean: 

We are looking for the value that is representative of the center of the data; thus the mean would be the best measure because of how varied the set is. Normally, when a data set is varied the mean is normally the best measure of center.

In order to answer this question correctly, we need to solve for the mean, median, and mode of this data set. 

To begin, let's sort the data from least to greatest:

Now that our data is ordered from least to greatest, we can solve for the median:

Remember, the median is the middle most number when a data set is ordered from least to greatest.

The median for this data set is 

Next, we can look at our data set to determine the mode:

The mode for this set is 

Remember, the mode is the number in a set that appears most often. 

Finally, we can solve for the mean:

Remember, the mean of a data set is the average of the numbers in a data set. 

The mean for this data set is 

Now that we've done our calculation we should have:

Median: 

Mode: 

Mean: 

We are looking for the value that is representative of the center of the data. In this data set we have an outlier, which means the mean is not going to be the best measure of center. Also, the mode is the lowest value in our set; thus, the median is the best measure of center. 

In order to answer this question correctly, we need to solve for the range and the interquartile range. 

To begin, let's sort the data from least to greatest:

Next, we can solve for the range. Remember, the range of a data set is the difference between the highest value and the lowest value in the set. 

The range for this data set is 

Now, we can solve for the interquartile range. Remember, the interquartile range is the difference between the upper quartile median and the lower quartile median. This means that we need to first calculate these two values. In order to do this, we need to split the data set into quartiles. 

First, we will find the median:

We will then use the median to split the data in half. Next,  we must find the median of the first half—or lower quartile—and then the median of the second half—or upper quartile:

Now we can solve for the difference between the upper quartile median and the lower quartile median:

Now that we have completed these operations, we should have calculated the following values:

Range: 

Interquartile range: 

As you can see, solving for the interquartile range requires more steps because it takes into account more of the data points; thus, given our options, interquartile range is best to use when solving for variability. 

In order to answer this question correctly, we need to solve for the mean, median, and mode of this data set. 

To begin, let's sort the data from least to greatest:

Now that our data is ordered from least to greatest, we can solve for the median:

Remember, the median is the middle most number when a data set is ordered from least to greatest.

The median for this data set is 

Next, we can look at our data set to determine the mode:

The mode for this set is 

Remember, the mode is the number in a set that appears most often. 

Finally, we can solve for the mean:

Remember, the mean of a data set is the average of the numbers in a data set. 

The mean for this data set is 

Now that we've done our calculation we should have:

Median: 

Mode: 

Mean: 

We are looking for the value that is representative of the center of the data. In this data set we have an outlier, which means the mean is not going to be the best measure of center. Also, the mode is the lowest value in our set; thus, the median is the best measure of center. 

In order to answer this question correctly, we need to solve for the range and the interquartile range. 

To begin, let's sort the data from least to greatest:

Next, we can solve for the range. Remember, the range of a data set is the difference between the highest value and the lowest value in the set. 

The range for this data set is 

Now, we can solve for the interquartile range. Remember, the interquartile range is the difference between the upper quartile median and the lower quartile median. This means that we need to first calculate these two values. In order to do this, we need to split the data set into quartiles. 

First, we will find the median:

We will then use the median to split the data in half. Next,  we must find the median of the first half—or lower quartile—and then the median of the second half—or upper quartile:

Now we can solve for the difference between the upper quartile median and the lower quartile median:

Now that we have completed these operations, we should have calculated the following values:

Range: 

Interquartile range: 

As you can see, solving for the interquartile range requires more steps because it takes into account more of the data points; thus, given our options, interquartile range is best to use when solving for variability. 

The first thing that we want to do when dividing decimals is to turn the divisor into a whole number. We do this by moving the decimal place to the right: 

If we move the decimal over one place in the divisor, we must also move the decimal over one place in the dividend:  

The new division problem should look as follows:  

*Notice how we've already placed the decimal in our answer. When we divide decimals, we place the decimal directly above the decimal in the dividend, but only after we've completed the first two steps of moving the decimal point in the divisor and dividend.

Now we can divide like normal:

Think: how many times can 76 go into 197

76 can go into 197 two times so we write a 2 over the 7 in the dividend:

Next, we multiply 2 and 76 and write that product underneath the 197 and subtract:

Now we bring down the 6 from the dividend to make the 45 into a 456.

Think: how many times can 76 go into 456?

76 can go into 465 six times so we write a 6 above the 6 in the dividend:

Next, we multiply 6 and 76 and write that product underneath the 456 and subtract:

We are left with no remainder and a final quotient of 2.6

The first thing that we want to do when dividing decimals is to turn the divisor into a whole number. In this case, the divisor is already a whole number so no change is needed. 

The division problem should look as follows:

*Notice how we've already placed the decimal in our answer. When we divide decimals, we place the decimal directly above the decimal in the dividend, but only after we've completed the first two steps of moving the decimal point in the divisor and dividend.

Now we can divide like normal:

Think: how many times can 12 go into 8

12 cannot go into 8 so we write a 0 over the 8 in the dividend:

Since 12 could not go into 8 we combine the ones place and tenths place and think of how many times 12 can go into 85. The number is split with the decimal but for multiplication's sake, we think of it as just an 85.

Think: how many times can 12 go into 85

12 can go into 85 seven times so we write a 7 above the 5 in the dividend:

Next, we multiply 12 and 7 and write that product underneath the 85 and subtract:

Now we bring down the 8 from the dividend to make the 1 into an 18.

Think: how many times can 12 go into 18?

12 can go into 18 one times so we write a 1 above the 8 in the dividend:

Next, we multiply 12 and 1 and write that product underneath the 18 and subtract:

Now we are left with 6 in our dividend and we cannot multiply 12 by anything to make a 6. We annex or add a zero to our dividend which we can carry down beside the 6 and it will now be a 60. We did no change the value of our dividend, we added a zero to make the number divisible by 12.

Think: how many times can 12 go into 60?

12 can go into 60 five times so we write a 5 above the 0 in the dividend:

Next, we multiply 12 and 5 and write that product underneath the 60 and subtract:

We are left with no remainder and a final quotient of 0.715

The first thing that we want to do when dividing decimals is to turn the divisor into a whole number. In this case, the divisor is already a whole number so no change is needed. 

The division problem should look as follows:

*Notice how we've already placed the decimal in our answer. When we divide decimals, we place the decimal directly above the decimal in the dividend, but only after we've completed the first two steps of moving the decimal point in the divisor and dividend.

Now we can divide like normal:

Think: how many times can 9 go into 8

9 cannot go into 8 so we write a 0 over the 8 in the dividend:

Since 9 could not go into 8 we combine the ones place and tenths place and think of how many times 9 can go into 87. The number is split with the decimal but for multiplication's sake, we think of it as just an 87.

Think: how many times can 9 go into 87

9 can go into 87 nine times so we write a 9 above the 7 in the dividend:

Next, we multiply 9 and 9 and write that product underneath the 87 and subtract:

Now we bring down the 3 from the dividend to make the 6 into an 63.

Think: how many times can 9 go into 63?

9 can go into 63 seven times so we write a 7 above the 3 in the dividend:

Next, we multiply 9 and 7 and write that product underneath the 63 and subtract:

We are left with no remainder and a final quotient of 0.97

The first thing that we want to do when dividing decimals is to turn the divisor into a whole number. We do this by moving the decimal place to the right: 

If we move the decimal over one place in the divisor, we must also move the decimal over one place in the dividend:  

The new division problem should look as follows:  

*Notice how we've already placed the decimal in our answer. When we divide decimals, we place the decimal directly above the decimal in the dividend, but only after we've completed the first two steps of moving the decimal point in the divisor and dividend.

Now we can divide like normal:

Think: how many times can 18 go into 45

18 can go into 45 two times so we write a 2 over the 5 in the dividend:

Next, we multiply 2 and 18 and write that product underneath the 45 and subtract:

Now 18 cannot be multiplied by a whole number to create a 9 so annex or add a zero to the dividend to create a number divisible by 18. We are not changing the value of the dividend by adding a zero. Bring that 0 down next to the 9 to create 90.

Think: how many times can 18 go into 90?

18 can go into 90 five times so we write a 5 above the 0 in the dividend:

Next, we multiply 5 and 18 and write that product underneath the 90 and subtract:

We are left with no remainder and a final quotient of 2.5

The first thing that we want to do when dividing decimals is to turn the divisor into a whole number. We do this by moving the decimal place to the right: 

If we move the decimal over one place in the divisor, we must also move the decimal over one place in the dividend:  

The new division problem should look as follows:  

*Notice how we've already placed the decimal in our answer. When we divide decimals, we place the decimal directly above the decimal in the dividend, but only after we've completed the first two steps of moving the decimal point in the divisor and dividend.

Now we can divide like normal:

Think: how many times can 52 go into 1

52 cannot go into 1 so we write a 0 over the 1 in the dividend:

We did not use the 1 in the hundreds place so now we bring in the 0 in the tens place and attempt to divide it by 52

Think: how many times can 52 go into 10

52 cannot go into 10 so we write a 0 over the 0 in the dividend:

We did not use the 10 so we now bring in the 1 from the ones place and attempt to divide it by 52

Think: how many times can 52 go into 101

52 can go into 101 one time so we write a 1 over the 1 in the dividend:

Next, we multiply 52 and 1 and write that product underneath the 101 and subtract:

Now we bring down the 4 from the dividend to make the 49 into a 494.

Think: how many times can 52 go into 494

52 can go into 494 nine times so we write a 9 over the 4 in the dividend:

Next, we multiply 52 and 9 and write that product underneath the 494 and subtract:

Now 52 cannot be multiplied by a whole number to create a 26 so annex or add a zero to the dividend to create a number divisible by 52. We are not changing the value of the dividend by adding a zero. Bring that 0 down next to the 26 to create 260.

Think: how many times can 52 go into 260?

52 can go into 260 five times so we write a 5 above the 0 in the dividend:

Next, we multiply 52 and 5 and write that product underneath the 260 and subtract:

We are left with no remainder and a final quotient of 1.95