Question

now that all of the lower class limits have been determined, we can find the upper class limits.
we must ensure that each value in the data set, which in this case only contains whole numbers, can belong to a class. this means the upper class limit for an earlier class will be one less than the lower class limit for the following class.
so, to find the upper class limit for the first class, subtract 1 from the lower class limit of the select class. that is, the following.
\boxed{} - 1 = 45
continue with this process until each upper class limit is determined.
\

$$\begin{tabular}{|c|c|c|} \\hline & lower class limit & upper class limit \\\\ \\hline first class & 30 & 45 \\\\ \\hline second class & 46 & \\boxed{} \\\\ \\hline third class & 62 & 77 \\\\ \\hline fourth class & 78 & \\boxed{} \\\\ \\hline fifth class & 94 & 109 \\\\ \\hline sixth class & 110 & \\boxed{} \\\\ \\hline seventh class & 126 & 141 \\\\ \\hline \\end{tabular}$$

Explanation:

Step1: Find the missing value in the equation

We have the equation \( x - 1 = 45 \). To find \( x \), we add 1 to both sides of the equation. So \( x = 45 + 1 = 46 \). This means we subtract 1 from the lower class limit of the second class to get the upper class limit of the first class.

Step2: Find the upper class limit of the second class

The lower class limit of the third class is 62. So the upper class limit of the second class is \( 62 - 1 = 61 \).

Step3: Find the upper class limit of the fourth class

The lower class limit of the fifth class is 94. So the upper class limit of the fourth class is \( 94 - 1 = 93 \).

Step4: Find the upper class limit of the sixth class

The lower class limit of the seventh class is 126. So the upper class limit of the sixth class is \( 126 - 1 = 125 \).

Answer:

The missing value in the first equation is 46. The upper class limit of the second class is 61, the fourth class is 93, and the sixth class is 125.