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:
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:
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:
d) The expected flight time is E(t)=122.5
The answer is 18/5 and simplify that and u get 3 3/5.
The answer to the question provided is 14
1.) 44
2.) 29
3.) 27
4.) 18
5.) 37
6.) 24
7.) 42
8.) 506
9.) 0
10.) -5
The linear model of this case takes the form:
y = a(x-b) + k
<span>The cost of having a package delivered has a base fee of $9.70
this is "k" >>>>> k=9.7 (fixed amount of fee)
THEN
</span>
<span>Every pound over 5 lbs cost an additional $0.46 per pound
that means: 0.46(x-5)
in other words, if the package weighs foe example 9 pounds, then 9-5=4, it will cost 0.46*4 for these 4 extra pounds
Finally we have the linear form of this: C = 0.46 (W - 5) + 9.7
</span>
