PLEAS HELPP MEEE:< RELATIONS AND FUNCTIONS
A. Let A = {2, 3, 4} and B = {6, 8, 10} and define a relation R from A to B as follows: For all (x, y) ∈ A x B,
(x, y) ∈ R means that yx is an integer.
a. State explicitly which ordered pairs are in A x B and which are in R.
b. Is 4 R 6? Is 4 R 8? Is 3 R 8? Is 2 R 10?
c. What are the domain and co-domain of R?
Solution:

B. Let E = {1, 2, 3} and F = {-2, -1, 0} and define a relation R from E to F as follows: For all (x, y) ∈ E x F,
(x, y) ∈ R means that x - y3 is an integer.
a. Which set of ordered pairs are in R.
b. Is 3 R 0? Is 1 R (-1)? Is 2 R (-1)? Is (3, -2) ∈ R?
c. What are the domain and co-domain of R?
Solution:

C. Let X = {2, 4, 5} and Y = {1, 2, 4, 6}. Which of the following determine functions from X to Y?

1. {(2, 6), (2, 2), (1, 1), (5, 4)}


2. {(2, 4), (4, 1)}


3. {(2, 4), (4, 1), (4, 2), (5, 6)}


4. {(2, 6), (4, 2), (5, 2)}


5. {(4, 2), (5, 4)}
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the answer   is 200

step-by-step explanation:

this is a problem about poisson probabilities.   i don't know if you've

already studied poisson distributions, or if you have a place to look

to read more about the subject.  

here's how you can see a poisson distribution coming.   when there are

a lot of events that may happen at any time (or any place), and when

you're counting how many of them happen at a particular time (or a

particular place), and each event is independent of the others, then

the result is a poisson distribution.   this is roughly true for the

chocolate chips; one chip being in a cookie has only a small effect on

the probability that another chip will be in the cookie.

once you know you have a poisson distribution, you know a lot.   there

is only one parameter to a poisson distribution, so if you know the

mean, you know everything.   the form of the distribution is

  probability of n chips = {[e^(-x)]*(x^n)} / (n! )