Answer:
The correct answer is:
a. M54.6, C79.51, C80.1
Explanation:
- M54.6 Pain in thoracic spine. It is a billable/specific ICD-10-CM code that can be used to indicate a diagnosis for reimbursement purposes. The 2020 edition of ICD-10-CM M54.
- C79.51: Secondary malignant neoplasm of bone, it is a billable/specific ICD-10-CM code that can be used to indicate a diagnosis for reimbursement purposes.
- G89. 3 is a billable/specific ICD-10-CM code that can be used to indicate a diagnosis for reimbursement purposes. The 2020 edition of ICD-10-CM G89.
Malignant neoplasm of anus, unspecified
Neoplasm related pain (acute) (chronic)
Pain in thoracic spine. M54. 6 is a billable/specific ICD-10-CM code that can be used to indicate a diagnosis for reimbursement purposes. The 2020 edition of ICD-10-CM M54.
Malignant (primary) neoplasm, unspecified
- C80. 1 is a billable/specific ICD-10-CM code that can be used to indicate a diagnosis for reimbursement purposes. The 2020 edition of ICD-10-CM C80.
Answer:
The solution code is written in Python 3 as below:
- outfile = open("greeting.txt", "w")
- outfile.write("Hello World")
- outfile.close()
Explanation:
To create a simple text file in Python, we can use Python built-in function, <em>open()</em>. There are two parameters needed for the open() function,
- the file name and
- a single keyword "w". "w" denote "write". This keyword will tell our program to create a file if the file doesn't exist.
The statement <em>open("greeting.txt", "w")</em> will create a text file named "<em>greeting.txt</em>" (Line 1)
To fill up the content in the greeting.txt, we use <em>write()</em> method. Just include the content string as the argument of the <em>write()</em> method. (Line 2)
At last, we use <em>close() </em>method to close the opened file,<em> outfile</em>. This will release the system resource from the<em> outfile.</em>
Answer:
Pseudo CODE
a)
n= Input “Enter 5 integer value”
b)
sum=0.0
For loop with i ranging from 0 - 5
Inside loop sum=n[i]+sum
Outside loop avg= sum/5
Print avg
c)
small=n[0] # assume the first number in the list is smallest
large= n[0] # assume the first number in the list is largest
For loop with i ranging from 0 - 5
Inside loop if n[i]<small #if any another number is smaller than small(variable)
Inside if Then small=n[i]
Inside loop if n[i]>large # if any another number is larger than large(variable)
Inside if then large=n[i]
Print small
Print large
d)
print avg
print small
print large
Answer:
C) /etc/updatedb.conf
Explanation:
The locate command actually uses the configuration file located at /etc/updated.conf.
Answer:
Here is the Python program:
d = {5:3, 4:1, 12:2}
val_of_max = d[max(d.keys())]
print(val_of_max)
Explanation:
The program works as follows:
So we have a dictionary named d which is not empty and has the following key-value pairs:
5:3
4:1
12:2
where 5 , 4 and 12 are the keys and 3, 1 and 2 are the values
As we can see that the largest key is 12. So in order to find the largest key we use max() method which returns the largest key in the dictionary and we also use keys() which returns a view object i.e. the key of dictionary. So
max(d.keys()) as a whole gives 12
Next d[max(d.keys())] returns the corresponding value of this largest key. The corresponding value is 2 so this entire statement gives 2.
val_of_max = d[max(d.keys())] Thus this complete statement gives 2 and assigns to the val_of_max variable.
Next print(val_of_max) displays 2 on the output screen.
The screenshot of program along with its output is attached.