the cafeteria use 9 2/4 gallons of milk on Monday and 15 1/4 gallons of milk on Tuesday how much more milk did they use on Tuesd
2 answers:
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First of all, since we have 36 characters available per spot (26 letters and 10 digits), and we have 6 spots, we have a total of
possible passwords.
Event A happens if the password starts with either a, e, i, o or u. If we fix the first character, we're left with 36 characters available for each of the remaining 5 spots, leading to a total of
possible passwords.
So, the probability of event A, computed as the ratio between "good" cases and all possible cases, is
Event B works exactly the same, since we're fixing the last spot, leaving 36 characters available for each of the first 5 spots. So, we have
As for the intersection, we want the first character to be a vowel, and the last character to be an even digits. There are 25 passwords satisfying this request:
Where x can be any of the 36 characters.
So, we have 25 cases with 4 vacant slots, leading to a probability of
Finally, you can compute the probability of the union using the formula
Since we already computed all these quantities.
Answer:
The probability that the restaurant can accommodate all the customers who do show up is 0.3564.
Step-by-step explanation:
The information provided are:
- At 7:00 pm the restaurant can seat 50 parties, but takes reservations for 53.
- If the probability of a party not showing up is 0.04.
- Assuming independence.
Let <em>X</em> denote the number of parties that showed up.
The random variable X follows a Binomial distribution with parameters <em>n</em> = 53 and <em>p</em> = 0.96.
As there are only 50 sets available, the restaurant can accommodate all the customers who do show up if and only if 50 or less customers showed up.
Compute the probability that the restaurant can accommodate all the customers who do show up as follows:
Thus, the probability that the restaurant can accommodate all the customers who do show up is 0.3564.