Answer:
Let number of cans collected by Shane =S
Number of cans collected by Abha=S+178
As, it is given that ,Abha collected 178 more cans than Shane did.
Also, the statement about inequality is
Both Shane and Abha have to collect no less than 2000 cans for recycling.
Writing the statement in terms of inequality
S +S+178 ≥ 2000
2 S ≥ 2000 - 178
2 S ≥ 1822
Dividing both sides by 2, we get
S≥ 911
So, Shane must have collected At least 911 cans.
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

1)

Each marker cost 59 cents
2)

The old Big Almond Bag weighed 15 ounces. The new Big Almond Bag weighs 18 ounces.
Answer:
3,474.4
Step-by-step explanation:
4,250-25%=3187.5
3187.5+9%=3474.4