The answers would be (0,10) , (20,0) , and (100,100)
Answer:
a) 7.79%
b) 67.03%
c) Cumulative Distribution Function

Step-by-step explanation:
We are given the following in the question:

where x is the duration of a call, in minutes.
a) P( calls last between 2 and 3 minutes)
![=\displaystyle\int^3_2 p(x)~ dx\\\\= \displaystyle\int^3_20.1e^{-0.1x}~dx\\\\=\Big[-e^{-0.1x}\Big]^3_2\\\\=-\Big[e^{-0.3}-e^{-0.2}\Big]\\\\= 0.0779\\=7.79\%](https://tex.z-dn.net/?f=%3D%5Cdisplaystyle%5Cint%5E3_2%20p%28x%29~%20dx%5C%5C%5C%5C%3D%20%5Cdisplaystyle%5Cint%5E3_20.1e%5E%7B-0.1x%7D~dx%5C%5C%5C%5C%3D%5CBig%5B-e%5E%7B-0.1x%7D%5CBig%5D%5E3_2%5C%5C%5C%5C%3D-%5CBig%5Be%5E%7B-0.3%7D-e%5E%7B-0.2%7D%5CBig%5D%5C%5C%5C%5C%3D%200.0779%5C%5C%3D7.79%5C%25)
b) P(calls last 4 minutes or more)
![=\displaystyle\int^{\infty}_4 p(x)~ dx\\\\= \displaystyle\int^{\infty}_40.1e^{-0.1x}~dx\\\\=\Big[-e^{-0.1x}\Big]^{\infty}_4\\\\=-\Big[e^{\infty}-e^{-0.4}\Big]\\\\=-(0- 0.6703)\\= 0.6703\\=67.03\%](https://tex.z-dn.net/?f=%3D%5Cdisplaystyle%5Cint%5E%7B%5Cinfty%7D_4%20p%28x%29~%20dx%5C%5C%5C%5C%3D%20%5Cdisplaystyle%5Cint%5E%7B%5Cinfty%7D_40.1e%5E%7B-0.1x%7D~dx%5C%5C%5C%5C%3D%5CBig%5B-e%5E%7B-0.1x%7D%5CBig%5D%5E%7B%5Cinfty%7D_4%5C%5C%5C%5C%3D-%5CBig%5Be%5E%7B%5Cinfty%7D-e%5E%7B-0.4%7D%5CBig%5D%5C%5C%5C%5C%3D-%280-%090.6703%29%5C%5C%3D%200.6703%5C%5C%3D67.03%5C%25)
c) cumulative distribution function

- 47 → d
using the ' product rule '
(uv) = uv' + vu' = (3 × - 4) + (7 × - 5 ) = - 12 - 35 = - 47
Answer:
Step-by-step explanation:
We take two points (0,-3) and (4,0)
eq. of line thro' two points is y-y1=m(x-x1)
m=(y2-y1)/(x2-x1)=(0+3)/(4-0)=3/4
Reqd. eq . is y-(-3)=3/4(x-0)
or y=3/4(x)-3
Answer:
Can you add the picture please???
Step-by-step explanation: