2. A taxi company has three cabs. Calls come in to the dispatcher at times of a Poisson process with rate 2 per hour. Suppose that each requires an exponential amount of time with mean 20 minutes, and that callers will hang up if they hear there are no cabs available. (a) What is the probability all three cabs are busy when a call comes in

0.406

Step-by-step explanation:

Let us make the number of cabs available to be x.

Number of calls is distributed as Poisson(\lambda =2) and

the availability of cab is distributed as Exponential(\theta =3).

Therefore, the probability all three cabs are busy when a call comes in is given by;

P(X=0)=\frac{e^{-\lambda}\lambda^x}{x!}\theta e^{-\theta x}=\frac{e^{-2}2^0}{0!}3e^{-3\times 0}= 0.406

answer: 42

use a calculator to find that 845/73 = 11.575

the whole part of the answer is 11. multiply this by 73 to get 11*73 = 803.

subtract that result from 845 to get 845 - 803 = 42

step-by-step explanation:

given that ab is divided into equal parts as

ac, cd, de, ef, fg, gh, hi, ib.

from this we find that f is the midpoint of ab.

so using mid point formula we find

f = (3,7)

similarly considering fb, we get

h = mid point of fb = ((6,6) and g, mid point of fh is

g = (9/2, 13/2)

i is mid point of hb

so i =(15/2,11/2)