Let's say you want to compute the probability

where

converges in distribution to

, and

follows a normal distribution. The normal approximation (without the continuity correction) basically involves choosing

such that its mean and variance are the same as those for

.
Example: If

is binomially distributed with

and

, then

has mean

and variance

. So you can approximate a probability in terms of

with a probability in terms of

:

where

follows the standard normal distribution.
Answer: i have no idea, but i wish you the best!!!! ;)
Step-by-step explanation:
Answer:
What? Hey no no no yes what no no yes what yes yes
Ab = (7x-6) → b = (7x-6)/a (1)
bc = (12-2x) → b = (12-2x)/c (2)
Since (1) = (2) → (7x-6)/a = (12-2x)/c OR a/c = (7x-6)/(12-2x) (3)
Multiply both numerators of (3) by the SQUARE of their respective denominators;
(a*c²)/c = (7x-6)(12-2x)²/(12-2x)
Now simplify:
ac = (7x-6)(12x-2x) or ac = -14x² + 96x - 72