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.
The best option is A) d - 90.
Expressions has no equal sign nor an inequality symbol. An expression has only standard or normal numbers, with variables and some basic mathematical operations.
3 more. The color doesn't matter at all. It's all about the amount.
I assume that the numbers are supposed to be written as 7.14 x 10^10 for December and 3.21 x 10^10. The total mail delivered is the sum of these two numbers.
(7.14 x 10^10) + (3.21 x 10^10) = 1.035 x 10^11
Answer:
Step-by-step explanation:
c^2 = a^2 + b^2
4.5^2 + 9^2 = 101.25
c = sqrt 101.25 = 10.06 = 10.1 meters