Answer:
0.281 ; 0.864
Step-by-step explanation:
1.)
p decreases as n increases (n increases, p--> 0)
λ ≥ 0
P(x = x) = (e^-λ * λ^x) ÷ x!
λ for Ford = 1.27
B) zero problems :
P(x = 0) :
(e^-1.27 * 1.27^0) ÷ 0! = 0.281
P(x = 0) = 0.281
Two or fewer problems :
P(x ≤ 2) = p(x = 0) + p(x = 1) + p(x = 2)
P(x ≤ 2) = 0.281 + 0.357 + 0.226
P(x ≤ 2) = 0.864
D.) It allows for early detection and thus adequate curtailment of the problem at an a stage in which it seems easier to curb.
Answer:6 0
Step-by-step explanation:
Multiply 12 by 37%.
12 × 0.37 = 4.44
B) at least 4
It could also be c... 4.44 is less than 8. But B is the closest answer to 4.44.
I’m not 100% but I think it could be the last table
