Answer:
a) Attached
b) P=0.60
c) P=0.80
d) The expected flight time is E(t)=122.5
Step-by-step explanation:
The distribution is uniform between 1 hour and 50 minutes (110 min) and 135 min.
The height of the probability function will be:
![h=\frac{1}{Max-Min}=\frac{1}{135-110} =\frac{1}{25}](https://tex.z-dn.net/?f=h%3D%5Cfrac%7B1%7D%7BMax-Min%7D%3D%5Cfrac%7B1%7D%7B135-110%7D%20%3D%5Cfrac%7B1%7D%7B25%7D)
Then the probability distribution can be defined as:
![f(t)=\frac{1}{25}=0.04 \,\,\,\,\\\\t\in[110,135]](https://tex.z-dn.net/?f=f%28t%29%3D%5Cfrac%7B1%7D%7B25%7D%3D0.04%20%5C%2C%5C%2C%5C%2C%5C%2C%5C%5C%5C%5Ct%5Cin%5B110%2C135%5D)
b) No more than 5 minutes late means the flight time is 125 or less.
The probability of having a flight time of 125 or less is P=0.60:
![F(T](https://tex.z-dn.net/?f=F%28T%3Ct%29%3D0.04%28t-min%29%5C%5C%5C%5CF%28T%3C125%29%3D0.04%2A%28125-110%29%3D0.04%2A15%3D0.60)
c) More than 10 minutes late means 130 minutes or more
The probability of having a flight time of 130 or more is P=0.80:
![F(T>t)=1-0.04(t-110)\\\\F(T>130)=1-0.04*(130-110)=1-0.04*20=1-0.8=0.2](https://tex.z-dn.net/?f=F%28T%3Et%29%3D1-0.04%28t-110%29%5C%5C%5C%5CF%28T%3E130%29%3D1-0.04%2A%28130-110%29%3D1-0.04%2A20%3D1-0.8%3D0.2)
d) The expected flight time is E(t)=122.5
![E(t)=\frac{1}{2}(max+min)= \frac{1}{2}(135+110)=\frac{1}{2}*245=122.5](https://tex.z-dn.net/?f=E%28t%29%3D%5Cfrac%7B1%7D%7B2%7D%28max%2Bmin%29%3D%20%5Cfrac%7B1%7D%7B2%7D%28135%2B110%29%3D%5Cfrac%7B1%7D%7B2%7D%2A245%3D122.5)
Answer:
1. 2x + 30° = 180°
2x = 150°
x = 75°
2. (2x - 24)° + (x + 12)° + (x - 8)° = 180°
4x - 20 = 180°
4x = 200
x = 50°
Answer:
10000 times 3 times 0.04 = 1200
Since he went 168 miles in 4 hours, then his speed is 42 mph. Then, 504 divided by 42 is 12. It will take 12 hours at a steady pace of 42 mph to reach 504 miles.