Which intervals can you use to make a frequency table for the data set? 95 103 110 86 72 77 97 118 83 101 105 79 78 95 93

To choose an interval, one should always first find the range i.e. highest and the lowest value difference.
Here, from the data set : 95 103 110 86 72 77 97 118 83 101 105 79 78 95 93

The  highest number is  118 and the smallest number is 72, and the difference between the two is (118 - 72) = 46 

And, divide 46 by 5, we get, 9.2 which is approx = 9.

Now, the lower class boundary will be starting with 72, then,  72 + 9 = 81, then, 81+ 9 = 90. and so on.

Similarly, the upper-class boundary will be one less than the second lowest boundary. i.e. 81-1 = 80, then, 90-1 = 89, and so on. 

LCB              UCB
72                   80
81                   89
90                   98
99                   107
108                  116
117                    125
126                   134

I’m going to infer you said parallel.solve for yy= -1/2x+7/4lines that are parallel have the same slope, different y intercept.so we know m is -1/2then from there solve for y intercept.