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Luda [366]
2 years ago
8

terms found in provider documentation (e.g., difficult, extensive, or unusual) would support adding modifier to a cpt code.

Computers and Technology
1 answer:
Oksi-84 [34.3K]2 years ago
3 0

The terms found in provider documentation (e.g., difficult, extensive, or unusual) would support adding modifier -22 to a cpt code.

<h3>What are cpt code?</h3>

Current Procedural Terminology (CPT) is known to be a kind of  a medical code set that is known to be used to report any form of medical, surgical, and diagnostic work and services to firms such as physicians, health insurance firms and accreditation firms.

Therefore, Note that the terms found in provider documentation (e.g., difficult, extensive, or unusual) would support adding modifier -22 to a cpt code.

Learn more about documentation from

brainly.com/question/968907

#SPJ1

Terms found in provider documentation (e.g., difficult, extensive, or unusual) would support adding modifier __________ to a CPT code.

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Which of the following shows a list of Big-Oh running times in order from slowest to fastest?
Rufina [12.5K]

Answer:

O(N!), O(2N), O(N2), O(N), O(logN)

Explanation:

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5 0
3 years ago
Read 2 more answers
Write a generator function named count_seq that doesn't take any parameters and generates a sequence that starts like this: 2, 1
-BARSIC- [3]

Answer:

#required generator function

def count_seq():

   #starting with number 2. using string instead of integers for easy manipulation

   n='2'

   #looping indefinitely

   while True:

       #yielding the integer value of current n

       yield int(n)

       #initializing an empty string

       next_value=''

       #looping until n is an empty string

       while len(n)>0:

           #extracting first digit (as char)

           first=n[0]

           #consecutive count of this digit

           count=0

           #looping as long as n is non empty and first digit of n is same as first

           while len(n)>0 and n[0]==first:

               #incrementing count

               count+=1

               #removing first digit from n

               n=n[1:]

           #now appending count and first digit to next_value

           next_value+='{}{}'.format(count,first)

       #replacing n with next_value

       n=next_value

#testing, remove if you don't need this

if __name__ == '__main__':

   #creating a generator from count_seq()

   gen=count_seq()

   #looping for 10 times, printing next value

   for i in range(10):

       print(next(gen))

Explanation:

6 0
3 years ago
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