Following ungrouped data of Sales of a company (in millions) is available 52 33 70 95 57 61
57 64 54 94 38 61
50 39 94 63 59 31
68 88 93 48 82 82
74 70 92 76 98 91
32 33 31 75 54 48
36 64 63 66 92 98
36 54 71 86 84 55

91 34 64 67 89 78
97 92 53 56 68 55
93 42 51 77 36 93
44 66 63 33 68 79
83 53 86 76 35 40
55 41 36 39 42 96
60 53 38 51 95 56
48 69 49 33 95 37
83 62 96 34 85 32
39 59 77 62 35 34
54 89 36 45 83 34
39 61 88 86 55 33
69 54 30 38 79 77
95 34 38 91 80 90
88 45 95 71 80 43
61 40 31 61 58 53
91 63 60 94 98 53
50 34 75 74 90 98

1. Make the frequency distribution table with appropiate class interval, frequency,
cumulative frequency.
1. Calculate the mean, median, mode and quartiles of grouped data.
2. Calculate the standard deviation, variance and range
3. Make the following diagrams: Histogram, Frequency Polygon, Olive

one that is bigger than 3 1/2

∠1≅∠2 by the alternate exterior angles theorem.

step-by-step explanation:

given, a ∥ b and ∠1 ≅ ∠3 .we have to prove that e ∥ f

we know that ∠1≅∠3 and that a || b because they are given. we see that by the alternate exterior angles theorem. therefore, ∠2≅∠3 by the transitive property. so, we can conclude that e || f by the converse alternate exterior angles theorem.

we have to fill the missing statement.

transitivity property states that if a = b and b = c, then a = c.

now, given ∠1≅∠3 and by transitivity property ∠2≅∠3 .

hence, to apply transitivity property one angle must be common which is not in result after applying this property which is ∠1.

the only options in which ∠1 is present are ∠1 and ∠2, ∠1 and ∠4

∠1 and ∠4 is not possible ∵ after applying transitivity we didn't get ∠4.

hence, the missing statement is ∠1≅∠2.

so, ∠1≅∠2 by the alternate exterior angles theorem.