A car salesman sold three hybrid cars last week. The cars come in red, white, and black. A sale is listed as RWB if a red car was sold first, a white was sold second, and a black car was sold third.

A list of all possible outcomes is shown. Each outcome has equal probability.

{RRR, RRW, RRB, RWR, RWW, RWB, RBR, RBW, RBB, WWW, WWR, WWB, WRR,

WRB, WRW. WBR, WBB, WBW, BBB, BBR, BBW, BWR, BWW, BWB, BRW, BRR, BRB)

What is the probability that the cars sold last week included a white car or the first car sold was red?

23 / 27

Step-by-step explanation:

Total possible outcomes :

{RRR, RRW, RRB, RWR, RWW, RWB, RBR, RBW, RBB, WWW, WWR, WWB, WRR, WRB, WRW. WBR, WBB, WBW, BBB, BBR, BBW, BWR, BWW, BWB, BRW, BRR, BRB) = 27

Required outcome : 23

{RRR, RRW, RRB, RWR, RWW, RWB, RBR, RBW, RBB, WWW, WWR, WWB, WRR, WRB, WRW. WBR, WBB, WBW, BBW, BWR, BWW, BWB, BRW) = 23

Required outcome / Total possible outcomes

= 23 / 27

∠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.

y=20x

: )

hope this