This requires the Poisson distribution, where
area = 5-acres
and mean number of field mice = 12 (in 5-acres of field)
therefore
lambda=12 (mean, given)
and the probability of k mice in the 5-acre field is given by the Poisson distribution as
P(X=k)=lambda^k * e^(-lambda) / k! ..............(1)
To find the probability of having LESS than 7 field mice, we add the probabilities of 0 to 6, which is
P(X<7)=P(X=0)+P(X=1)+...+P(X=6)
evaluating with equation (1) for X=0 to 6, we get:
0 0.0000061 0.0000742 0.0004423 0.0017704 0.0053095 0.0127416 0.025481Total = 0.045822
Answer: The probability that fewer than 7 field mice are found in the 5-acre field is 0.0458.
Answer:
This is nonlinear.
Step-by-step explanation:
Answer:
x + y < 20
5x + 10y ≥ 100
Step-by-step explanation:
The ski club wants at least $100
Student pays $5
Chaperone pays $10
Maximum number of seats = 20
Assumption:
Assumed number of students = x
Assumed number of chaperone = y
Equation:
Maximum number of seats = 20
⇒ x + y < 20
The club wants $100
⇒ 5x + 10y ≥ 100
Answer:
11
Step-by-step explanation:
50-6=44
44/4=11
Step-by-step explanation:
p = 0.1, q = 0.9, n = 7
a) Use complementary probability.
P(at least 1) = 1 − P(0)
P(at least 1) = 1 − (0.9)⁷
P(at least 1) = 0.522
b) Use binomial probability.
P = nCr pʳ qⁿ⁻ʳ
P(3) = ₇C₃ (0.1)³ (0.9)⁴
P(3) = 0.023
