Answer:
D.
Step-by-step explanation:
When a line is perpendicular to another, their slopes will be opposite reciprocal. For example, 1 would be -1, -3 would be
, and
would be -5. The equation is written in slope-intercept form:
![y=mx+b](https://tex.z-dn.net/?f=y%3Dmx%2Bb)
m is the slope, so find the equation with the opposite reciprocal of 3,
:
is the only option with the correct slope, so the answer is D.
GCF of 84 and 98 would be = to 14.
EDIT: Proofed it to be correct :)! You're welcome.
Answer:
y = -11
Step-by-step explanation:
its finaly answer , if any qeustion reply me again
Let's say you want to compute the probability
![\mathbb P(a\le X\le b)](https://tex.z-dn.net/?f=%5Cmathbb%20P%28a%5Cle%20X%5Cle%20b%29)
where
![X](https://tex.z-dn.net/?f=X)
converges in distribution to
![Y](https://tex.z-dn.net/?f=Y)
, and
![Y](https://tex.z-dn.net/?f=Y)
follows a normal distribution. The normal approximation (without the continuity correction) basically involves choosing
![Y](https://tex.z-dn.net/?f=Y)
such that its mean and variance are the same as those for
![X](https://tex.z-dn.net/?f=X)
.
Example: If
![X](https://tex.z-dn.net/?f=X)
is binomially distributed with
![n=100](https://tex.z-dn.net/?f=n%3D100)
and
![p=0.1](https://tex.z-dn.net/?f=p%3D0.1)
, then
![X](https://tex.z-dn.net/?f=X)
has mean
![np=10](https://tex.z-dn.net/?f=np%3D10)
and variance
![np(1-p)=9](https://tex.z-dn.net/?f=np%281-p%29%3D9)
. So you can approximate a probability in terms of
![X](https://tex.z-dn.net/?f=X)
with a probability in terms of
![Y](https://tex.z-dn.net/?f=Y)
:
![\mathbb P(a\le X\le b)\approx\mathbb P(a\le Y\le b)=\mathbb P\left(\dfrac{a-10}3\le\dfrac{Y-10}3\le\dfrac{b-10}3\right)=\mathbb P(a^*\le Z\le b^*)](https://tex.z-dn.net/?f=%5Cmathbb%20P%28a%5Cle%20X%5Cle%20b%29%5Capprox%5Cmathbb%20P%28a%5Cle%20Y%5Cle%20b%29%3D%5Cmathbb%20P%5Cleft%28%5Cdfrac%7Ba-10%7D3%5Cle%5Cdfrac%7BY-10%7D3%5Cle%5Cdfrac%7Bb-10%7D3%5Cright%29%3D%5Cmathbb%20P%28a%5E%2A%5Cle%20Z%5Cle%20b%5E%2A%29)
where
![Z](https://tex.z-dn.net/?f=Z)
follows the standard normal distribution.