Trip G at 1000 miles earning 5 points a mile plus an extra 20 for every 500
Answer:
The regular price of the ticket was $12.
Step-by-step explanation:
Let us call
the regular price of the tickets, and if she had bought tickets at this price, the total cost would have been
.
But Holly got $4 off the regular ticket price, that means one ticket cost her
and if she bought 23 tickets it cost her

and we are told this equals $184; therefore we have

now we solve this equation the following way:





Thus, the regular price was $12.
Answer:
1) The probability that ten students in a class have different birthdays is 0.883.
2) The probability that among ten students in a class, at least two of them share a birthday is 0.002.
Step-by-step explanation:
Given : Assume there are 365 days in a year.
To find : 1) What is the probability that ten students in a class have different birthdays?
2) What is the probability that among ten students in a class, at least two of them share a birthday?
Solution :

Total outcome = 365
1) Probability that ten students in a class have different birthdays is
The first student can have the birthday on any of the 365 days, the second one only 364/365 and so on...

The probability that ten students in a class have different birthdays is 0.883.
2) The probability that among ten students in a class, at least two of them share a birthday
P(2 born on same day) = 1- P( 2 not born on same day)
![\text{P(2 born on same day) }=1-[\frac{365}{365}\times \frac{364}{365}]](https://tex.z-dn.net/?f=%5Ctext%7BP%282%20born%20on%20same%20day%29%20%7D%3D1-%5B%5Cfrac%7B365%7D%7B365%7D%5Ctimes%20%5Cfrac%7B364%7D%7B365%7D%5D)
![\text{P(2 born on same day) }=1-[\frac{364}{365}]](https://tex.z-dn.net/?f=%5Ctext%7BP%282%20born%20on%20same%20day%29%20%7D%3D1-%5B%5Cfrac%7B364%7D%7B365%7D%5D)

The probability that among ten students in a class, at least two of them share a birthday is 0.002.
Answer:24%
Step-by-step explanation:
6 divided by 25 is .24