Answer:
a. P(x = 0 | λ = 1.2) = 0.301
b. P(x ≥ 8 | λ = 1.2) = 0.000
c. P(x > 5 | λ = 1.2) = 0.002
Step-by-step explanation:
If the number of defects per carton is Poisson distributed, with parameter 1.2 pens/carton, we can model the probability of k defects as:

a. What is the probability of selecting a carton and finding no defective pens?
This happens for k=0, so the probability is:

b. What is the probability of finding eight or more defective pens in a carton?
This can be calculated as one minus the probablity of having 7 or less defective pens.



c. Suppose a purchaser of these pens will quit buying from the company if a carton contains more than five defective pens. What is the probability that a carton contains more than five defective pens?
We can calculate this as we did the previous question, but for k=5.

Hey! We can’t see a picture of the problem.
Answer:
see explanation
Step-by-step explanation:
These are the terms of an arithmetic sequence with n th term
= a + (n - 1)d
where a is the first term and d the common difference
d = 25 - 20 = 30 - 25 = 5 and a = 20, hence
= 20 + 5(n - 1) = 20 + 5n - 5 = 5n + 15 ← n th term formula
Working is attached: 2x^3 - 4x^2 + 5x - 6 remainder 3
62.831 because the firs number that is different between the two is .8 and .3, and .8 is greater regardless of the numbers that follow afterward.