Answer:
0.087
Step-by-step explanation:
Given that there were 17 customers at 11:07, probability of having 20 customers in the restaurant at 11:12 am could be computed as:
= Probability of having 3 customers in that 5 minute period. For every minute period, the number of customers coming can be modeled as:
X₅ ~ Poisson (20 (5/60))
X₅ ~ Poisson (1.6667)
Formula for computing probabilities for Poisson is as follows:
P (X=ₓ) = ((<em>e</em>^(-λ)) λˣ)/ₓ!
P(X₅= 3) = ((<em>e</em>^(-λ)) λˣ)/ₓ! = (e^-1.6667)((1.6667²)/3!)
P(X₅= 3) = (2.718^(-1.6667))((2.78)/6)
P(X₅= 3) = (2.718^(-1.6667))0.46
P(X₅= 3) = 0.1889×0.46
P(X₅= 3) = 0.086894
P(X₅= 3) = 0.087
Therefore, the probability of having 20 customers in the restaurant at 11:12 am given that there were 17 customers at 11:07 am is 0.087.
Answer:
Step-by-step explanation:relax and claculate in ur mine
Answer: option d. x = 3π/2Solution:function y = sec(x)
1) y = 1 / cos(x)
2) When cos(x) = 0, 1 / cos(x) is not defined
3) cos(x) = 0 when x = π/2, 3π/2, 5π/2, 7π/2, ...
4) limit of sec(x) = lim of 1 / cos(x).
When x approaches π/2, 3π/2, 5π/2, 7π/2, ... the limit →+/- ∞.
So, x = π/2, x = 3π/2, x = 5π/2, ... are vertical asymptotes of sec(x).
Answer: 3π/2
The figures attached will help you to understand the graph and the existence of multiple asymptotes for y = sec(x).
-the equation is a quadratic
-the axis of symmetry is x=-7
-the graph has a relative maximum
U would simpiphy it to then get the awnser