Answer:
There are 1% probability that the last person gets to sit in their assigned seat
Step-by-step explanation:
The probability that the last person gets to sit in their assigned seat, is the same that the probability that not one sit in this seat.
If we use the Combinatorics theory, we know that are 100! possibilities to order the first 99 passenger in the 100 seats.
LIke we one the probability that not one sit in one of the seats, we need the fraction from the total number of possible combinations, of combination that exclude the assigned seat of the last passenger. In other words the amount of combination of 99 passengers in 99 seats: 99!
Now this number of combination of the 99 passenger in the 99 sets, divide for the total number of combination in the 100 setas, is the probability that not one sit in the assigned seat of the last passenger.
P = 99!/100! = 99!/ (100 * 99!) = 1/100
There are 1% probability that the last person gets to sit in their assigned seat
Answer:
2400 eggs
Step-by-step explanation:
12*200=2400
You'd start off by adding the 2 adults together (10.75 + 10.75) which would equal 21.50 dollars. Now you take the three kids together (5.50 +5.50+5.50) which would equal 16.50 dollars.
Now you just take the 16.50+21.50 and add those together---- 38 dollars
21) 3(2j-k)=108 It is asking you to set the equation equal to j
distribute the 3
6j-3k=108
move k to the other side by adding on both sides
6j=108+3k
divide everything by 6
j=18+0.5k
It will just be -7/12 because anything times 1 stays the same
