C is the incenter of isosceles triangle ABD with vertex angle ∠ABD. Does the following proof correctly justify that triangles ABC and DBC are congruent? It is given that C is the incenter of triangle ABD, so segment BC is an altitude of angle ABD.

Angles ABC and DBC are congruent according to the definition of an angle bisector.
Segments AB and DB are congruent by the definition of an isosceles triangle.
Triangles ABC and DBC share side BC, so it is congruent to itself by the reflexive property.
By the SAS postulate, triangles ABC and DBC are congruent.

There is an error in line 1; segment BC should be an angle bisector.
The proof is correct.
There is an error in line 3; segments AB and BC are congruent.
There is an error in line 5; the ASA Postulate should be used.

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! )

its electric no smoke

step-by-step explanation:

16

step-by-step explanation: