Recognizing the No-Solution Case Algebraically Consider the system of equations:
x2 + y2 = 8
x=3
Solving for one variable:
Work is shown to solve for the system of equations.
Why does the answer for y indicate that the system has
no real-number solutions?
1. Substitute:
(3)2 + y2 = 8
* The solution does not solve for x first.
O The values for y are square roots of negative
numbers.
The values for y have a positive and a negative
value.
O The values for y have square roots.
2. Simplify:
9 + y2 = 8
3. Isolate:
y²=-1
4. Solve fory:
y = -1
No solution

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! )

(-1,-0.5)

step-by-step explanation: