Step-by-step explanation:
Maybe it should be written as 4 9/10
In a mixed number
Answer: 1.25 times as far
Step-by-step explanation:
This problem requires you to find something called the unit rate.
Getting the unit rate for Car A is very simple, and can be found to be 25mi/gal.
To find the unit rate for Car B, simply look at the second row. Because the car can go 60 miles on 3 gallons of gas, divide both these numbers by 3 to get 20mi/gal.
Then, simply multiply both of these numbers by 11 to get 25 * 11 = 275, and 20 * 11 = 220.
Then, by doing 275/220, you get your answer: 1.25.
Hope it helps :) and let me know if you want me to elaborate.
Answer:
3y⁵√2
Step-by-step explanation:
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:
![P(X\leq 50)=1-P(X>50)\\=1-P(X=51)-P(X=52)-P(X=53)\\=1-[{53\choose 51}(0.96)^{51}(0.04)^{53-51}]-[{53\choose 52}(0.96)^{52}(0.04)^{53-52}]\\-[{53\choose 53}(0.96)^{53}(0.04)^{53-53}]\\=1-0.27492-0.25377-0.11491\\=0.3564](https://tex.z-dn.net/?f=P%28X%5Cleq%2050%29%3D1-P%28X%3E50%29%5C%5C%3D1-P%28X%3D51%29-P%28X%3D52%29-P%28X%3D53%29%5C%5C%3D1-%5B%7B53%5Cchoose%2051%7D%280.96%29%5E%7B51%7D%280.04%29%5E%7B53-51%7D%5D-%5B%7B53%5Cchoose%2052%7D%280.96%29%5E%7B52%7D%280.04%29%5E%7B53-52%7D%5D%5C%5C-%5B%7B53%5Cchoose%2053%7D%280.96%29%5E%7B53%7D%280.04%29%5E%7B53-53%7D%5D%5C%5C%3D1-0.27492-0.25377-0.11491%5C%5C%3D0.3564)
Thus, the probability that the restaurant can accommodate all the customers who do show up is 0.3564.