Answer:
(a) mean = 377.5ml. standard deviation = 4.33
(b) 0.333
(c) 384.25 ml
Step-by-step explanation:
(a)The mean:

The standard deviation:

(b)

(c)P(x > m) = 0.95



So the volumn of shampoo should be 384.25ml to exceed 95% of containers.
Assuming the order required is as n-> inf.
As n->inf, o(log(n+1)) -> o(log(n)) since the 1 is insignificant compared with n.
We can similarly drop the "1" as n-> inf, the expression becomes log(n^2+1) ->
log(n^2)=2log(n) which is still o(log(n)).
So yes, both are o(log(n)).
Note: you may have more offers of answers if you post similar questions in the computer and technology section.
Answer:
6,561
Step-by-step explanation:
(-9)^4 = -9*-9*-9*-9= 6,561
Let X be the number of burglaries in a week. X follows Poisson distribution with mean of 1.9
We have to find the probability that in a randomly selected week the number of burglaries is at least three.
P(X ≥ 3 ) = P(X =3) + P(X=4) + P(X=5) + ........
= 1 - P(X < 3)
= 1 - [ P(X=2) + P(X=1) + P(X=0)]
The Poisson probability at X=k is given by
P(X=k) = 
Using this formula probability of X=2,1,0 with mean = 1.9 is
P(X=2) = 
P(X=2) = 
P(X=2) = 0.2698
P(X=1) = 
P(X=1) = 
P(X=1) = 0.2841
P(X=0) = 
P(X=0) = 
P(X=0) = 0.1495
The probability that at least three will become
P(X ≥ 3 ) = 1 - [ P(X=2) + P(X=1) + P(X=0)]
= 1 - [0.2698 + 0.2841 + 0.1495]
= 1 - 0.7034
P(X ≥ 3 ) = 0.2966
The probability that in a randomly selected week the number of burglaries is at least three is 0.2966
You have to do rearranging formulae to find that out so basically solve for x