5 numbers :
mode = n median= n+1 mean = n+2 Range = n+3

find 2 sets of numbers

3,3,4,6,6

4,4,5,6,7

5,5,6,6,8

Step-by-step explanation:

mode = n

median= n+1

mean = n+2

Range = n+3

let sets:

a,b,c,d,e

mode = n, at least two of them are the same and equal to n

n,n,c,d,e

median belong to the set, in ascending order:

n,n,n+1,d,e

which means d,e ≥ n+1, d≠e

mean is n+2

5(n+2)=n+n+n+1+d+e

2n+9=d+e

Assume e is the maximum:

Range is n+3:

e-n=n+3

e=2n+3

d=6≥n+1

n≤5,

Set is n,n,n+1,6,n+3

n+1≤6≤n+3

3≤n≤5

3,3,4,6,6 bimodal

4,4,5,6,7

5,5,6,6,8 bimodal

∠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.