For each statement, write what would be assumed and what would be proven in a proof by contrapositive of the statement. Then write what would be assumed and what would be proven in a proof by contradiction of the statement. a. If x and y are a pair of consecutive integers, then x and y have opposite parity.
b. For all integers n, if n² is odd, then n is also odd.

a.)

Statement - If x and y are a pair of consecutive integers, then x and y have opposite parity.

Contra-positive statement - If x and y have same parity, then x and y can not be a pair of consecutive integers.

Assumed - x and y have same parity

To prove -  x and y can not be a pair of consecutive integers.

Contradiction statement - If x and y are a pair of consecutive integers and they have same parity then then it is not possible.

Assumed - x and y are a pair of consecutive integers and they have same parity

To prove - It is not possible

Proof - Let x and y are even numbers

⇒x = 2a , y = 2b

then , y - x = 2b - 2a = 2(b - a) is alteast 0 if x ≠y

Contradiction.

b.)

Statement - For all integers n, if n² is odd, then n is also odd.

Contrapositive statement - If n is even , then n² is even

assumed - n is even

To prove - n² is even

Contradiction - If n² is odd and n is even then it is not possible

Assumed - n² is odd and n is even

To prove - Our assumption is not true

Proof - As n is even

⇒n = 2a

⇒n² = (2a)² = 4a²

⇒n² is even

Contradiction.

there is no image or shapes attacted

step-by-step explanation:

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! )