Look at the right triangle ABC: Right triangle ABC has a right angle at B. Segment BD meets segment AC at a right angle.

A student made the following chart to prove that AB2 + BC2 = AC2:

Statement Justification
1. Triangle ABC is similar to triangle BDC 1. Angle ABC = Angle BDC and Angle BCA = Angle BCD
2. BC2 = AC ⋅ DC 2. BC ÷ DC = AC ÷ BC because triangle ABC is similar to triangle BDC
3. Triangle ABC is similar to triangle ABD 3. Angle ABC = Angle BAD and Angle BAC = Angle ABD
4. AB2 = AC ⋅ AD 4. AB ÷ AD = AC ÷ AB because triangle ABC is similar to triangle ADB
5. AB2 + BC2 = AC ⋅ AD + AC • DC = AC (AD + DC) 5. Adding Statement 1 and Statement 2
6. AB2 + BC2 = AC2 6. AD + DC = AC

Which justification is incorrect?
Justification 4
Justification 1
Justification 2
Justification 3

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! )

here is your mark brainliest