Answer:
-6 is your answer.
Step-by-step explanation:
(-2)(-2)(-2)= -6
Answer:
a) 3.6
b) 1.897
c)0.0273
d) 0.9727
Step-by-step explanation:
Rabies has a rare occurrence and we can assume that events are independent. So, X the count of rabies cases reported in a given week is a Poisson random variable with μ=3.6.
a)
The mean of a Poisson random variable X is μ.
mean=E(X)=μ=3.6.
b)
The standard deviation of a Poisson random variable X is √μ.
standard deviation=S.D(X)=√μ=√3.6=1.897.
c)
The probability for Poisson random variable X can be calculated as
P(X=x)=(e^-μ)(μ^x)/x!
where x=0,1,2,3,...
So,
P(no case of rabies)=P(X=0)=e^-3.6(3.6^0)/0!
P(no case of rabies)=P(X=0)=0.0273.
d)
P(at least one case of rabies)=P(X≥1)=1-P(X<1)=1-P(X=0)
P(at least one case of rabies)=1-0.0273=0.9727
Answer:
a. The probability that a customer purchase none of these items is 0.49
b. The probability that a customer purchase exactly 1 of these items would be of 0.28
Step-by-step explanation:
a. In order to calculate the probability that a customer purchase none of these items we would have to make the following:
let A represents suit
B represents shirt
C represents tie
P(A) = 0.22
P(B) = 0.30
P(C) = 0.28
P(A∩B) = 0.11
P(C∩B) = 0.10
P(A∩C) = 0.14
P(A∩B∩C) = 0.06
Therefore, the probability that a customer purchase none of these items we would have to calculate the following:
1 - P(A∪B∪C)
P(A∪B∪C) =P(A) + P(B) + P(C) − P(A ∩ B) − P(A ∩ C) − P(B ∩ C) + P(A ∩ B ∩ C)
= 0.22+0.28+0.30-0.11-0.10-0.14+0.06
= 0.51
Hence, 1 - P(A∪B∪C) = 1-0.51 = 0.49
The probability that a customer purchase none of these items is 0.49
b.To calculate the probability that a customer purchase exactly 1 of these items we would have to make the following calculation:
= P(A∪B∪C) - ( P(A∩B) +P(C∩B) +P(A∩C) - 2 P(A ∩ B ∩ C))
=0.51 -0.23 = 0.28
The probability that a customer purchase exactly 1 of these items would be of 0.28
Answer:
The LCM of 5, 6, and 9 is the product of all prime numbers on the left, i.e. LCM(5, 6, 9) by division method = 2 × 3 × 3 × 5 = 90.
