Answer:
0.1827
Step-by-step explanation:
Given mean of exponential distribution = 100
==> 1/χ = 100 ==> χ = 1/100 ==> χ = 0.01
PDF of χ , f(x) = χe^(-χx), x ≥ 0
===> f(x) = 0.01e^(-0.01x), x ≥ 0
Now we find the probability that the demand will exceed 170 cfs during the early afternoon on a randomly selected day
P(X>170) = <em>∞∫170 </em>f(x)dx
P(X>170) = <em>∞∫170 </em>0.01 e^(0.01x) dx
P(X>170) = [e^(-0.01x) / -0.01]^<em>∞ </em><u>base</u> 170
P(X>170) = -1 [e^-∞ - e^-0.01*170]
P(X>170) = e^-1.7
P(X>170) = 0.1827
The probability that the demand will exceed 170 cfs during the early afternoon on a randomly selected day is 0.1827
You don't have to look very far in either case. Both answers are "a".
1. Checking selection "a",
265 =? 35 +20*7 = 35 +140 = 175
a (7, 265) is NOT a solution
2. Checking selection "a",
18 =? 5*4 -2 = 20 -2 = 18
a (4, 18) IS a solution
Answer:
Step-by-step explanation:
5) 121/11 = 11 meals / day
6) 50/10 = 5 min/call
7) 91/7 = 13 books / week
8) 1275 ants/5 ant holes = 255 ants / ant hole