So, what is the total lenght of the pieces that wer cut off?
we multiply the lenght by the number:
4

*5=20+

=21

and we subtract this from the original pipe, that is
30

-21

we need to bring the two fractions to the same denominator (by multiplying the fraction art in the first by 3 and 4 the second):
30

-21

=9

so the correct answer is D!
Answer:
C
The "inverse operation" is just a way of saying "what do you do to isolate the variable". In this case, we isolate y, so we have to move all terms to the right side. To do that, we subtract 12 from each side to there will only be "y" on the left side.
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.

<span>the blank boxes are for you to plug in x=20 to prove its right.
so it would be
3(20)-4=2(20+8)
60-4=2(28)
56=56
so its true!</span>