There's no question there. You didn't attach the pic
Answer:
A
Step-by-step explanation:
Put brackets around the first two tems.
y = (x^2 - 8x) + 29
Take 1/2 coefficient of the linear term -8. Square that result. Add it inside the brackets.
1/2 (- 8) = - 4
(- 4)^2 = 16
y = (x^2 - 8x + 16) + 29
Subtract 16 outside the brackets.
y = (x^2 - 8x + 16) + 29 - 16
Do the subtraction
y = (x^2 - 8x + 16) + 13
Represent what is inside the brackets as a square.
y = ( x - 4)^2 + 13
The answer is A
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
Ax+3=7-bx
Ax=4-bx
A=4-bx/x
Ax+3=7-bx
Ax-4= -bx
Ax-4/x= -b
-Ax+4/x=b
b= -Ax+4/x
