Both of these conditions must be true in order for the assumption that the binomial distribution is approximately normal. In other words, if and
then we can use a normal distribution to get a good estimate of the binomial distribution. If either np or nq is smaller than 5, then a normal distribution wouldn't be a good model to use.
side note: q = 1-p is the complement of probability p
Answer:
<h2>DOMAIN: The set of all real numbers except 0</h2>Step-by-step explanation:
We have a faction.
We know: the denominator must be different from 0. Therefore
=======================================================
Explanation:
Extend segment MN such that it intersects side ST. Mark the intersection as point A. See the diagram below.
We're given that angle MNT is 72 degrees. The angle TNA is equal to 180-(angle MNT) = 180 - 72 = 108 degrees, since angles MNT and TNA add to 180.
For now, focus entirely on triangle TNA. We see from the diagram that T = 34 and we just found that N = 108. Let's find angle A
A+N+T = 180
A+108+34 = 180
A+142 = 180
A = 180-142
A = 38
So angle NAT is 38 degrees.
----------------------------
Since segment MA is an extension of MN, and because MN || SQ, this means MA is also parallel to SQ.
We found at the conclusion of the last section that angle NAT was 38 degrees. Angles QST and NAT are corresponding angles. They are congruent since MA || SQ. This makes angle QST to also be 38 degrees
----------------------------
The angles QSR and QST are a linear pair, so they are supplementary
(angle QSR) + (angle QST) = 180
angle QSR = 180 - (angle QST)
angle QSR = 180 - 38
angle QSR = 142 degrees
