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:
2
Step-by-step explanation:
Answer:
how did you get those warning signs in your question
Step-by-step explanation:
Answer:
3
√
2
Explanation:
First put change the words into an equation:
√
3
×
√
6
Now you can multiply them together as you would normally multiply:
√
3
×
√
6
=
√
18
Now let's prime factor 18 and see if there are any squares that we can take out of it to simplify. All we have to see is if there are 2 numbers that are the same:
18
/ \
6
3
/ \
2
3
As you can see, we have a square:
3
×
3
=
9
So take
√
9
out of
√
18
. You should have:
√
9
√
2
But since
√
9
=
3
we can simplify further to make:
√
9
√
2
→
3
√
2
Step-by-step explanation:
The answer is: [D]: " <span>(1, 0, 0), (0, –2, 0), (0, 0, 6) " .
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