In a state lottery, you must correctly select 5 numbers (in any order) out of 40 to win the top prize. 1. how many ways can 5 numbers be chosen from 40 numbers? 2. you purchase one lottery ticket. what is the probability that you will win the top prize? 3. if the order of the numbers mattered, will the probability of winning increase or decrease? why?

If the number has 5 numbers in any order (different ones)

The number of possible combinations is equal to product of the number of options that we have for each of the 5 numbers.

For the first one we have 40 numbers to choice.

for the second one we have 39, because we already took one.

for the third one we have 38

for the fourth we have 37

for the fifth we have 36

Then the number of combinations is:

40*37*38*37*36 = 78960960

and because the order does not matter, we need to divide by the posible permutations, for a 5 digit number the number of permutations is:

5 options for the first digit.

4 for the second.

3 for the third.

and so on:

P = 5*4*3*2*1 = 5!

so the number of combinations is: c = 78960960/5! = 658008

This means that if you buy a ticket, your probability of winning is 1 out of 658008, or p =  1/658008.

And if the order actually does matter, we use the previous number, with only 1 ticket the probability is 1 out of 78960960 or p = 1/78960960 which is a lot smaller.

idk

step-by-step explanation:

∠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.