In their never-ending quest to siphon the money from your wallet, the carnival operators have come up with a new game: Super-Mega-Chuck-A-Luck! Here you roll seven dice, with payoffs as follows: A grand prize of $5,000 if you roll seven of a kind $500 if you roll six of a kind $50 if you roll five of a kind What is the expected value of Super-Mega-Chuck-A-Luck?

The expected value is $0.035

Step-by-step explanation:

expected of outcome for each dice = 1,

each dice has 6 sides ⇒ total number of possible outcome per dice = 6

seven of a kind = $5,000, six of a kind = $500, five of a kind = $50

Probability = (expected outcome ÷ total number of possible outcome) ^ number of dice

Pr =

seven of a kind ⇒ all 7 dice have the same number

Pr (seven of a kind) = = 0.00000357

six of a kind ⇒ all 6 dice have the same number

Pr (six of a kind) = = 0.0000214

five of a kind ⇒ all 5 dice have the same number

Pr (five of a kind) = = 0.000129

EV = ∑P() *

EV = 5,000 * 0.00000357 + 500 * 0.0000214 + 50 * 0.000129

EV = 0.01785 + 0.0107 + 0.00645

EV = $0.035

The chances of winning is less than 1%. This is a very bad risk to take, do not let the carnival operators siphon the money in your wallet

∠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.

b and c

step-by-step explanation: