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:
neither
Step-by-step explanation:
the problem shows that it is neither
-y = x + 3 . . . . this is the sides for the double lines
2y + 24 = x + 9 . . . . this is the sides for the single lines
To solve you need to substitute one equation into the other
-y = x + 3
y = -x - 3 . . . place this equation into the other equation where (Y) is
2y + 24 = x + 9
2 ( -x - 3) + 4 = x +9
-2x - 6 + 24 = x + 9
-2x +18= x+9
-3x = -9
x = 3 . . . substitute this answer into the equation and solve for (Y)
-y = x + 3
- y = 3 +3
-y = 3 + 3
y= -6
Answer:
x = all real numbers.
Step-by-step explanation:
3 x minus 8 = negative x + 4 (x minus 2)
3x - 8 = -x + 4(x - 2)
3x - 8 = -x + 4x - 8
3x - 8 = 3x - 8
3x - 3x = -8 + 8
0 = 0
Since the result is a true statement, but 0 = 0, x is equivalent to all real numbers.
Hope this helps!