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.
You can try by plugging each ordered pair and seeing if the equation comes out true.
(5, 1)
2 * 5 - 1 = 9
That's correct so C is the correct answer.
Answer:
Step-by-step explanation:
The negative in front of the parenthesis changes the signs of the numbers and variables inside the parenthesis. We can rewrite as
4 - x - 1 = 6 and
3 - x = 6 so
- x = 3 and
x = -3
Answer:
the second number is 4
Step-by-step explanation:
13+x=17
x=4