The argument for the function would be answer "D".
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.
The code that examines all the strings in the input source and determines how long the longest string (or strings are) is the following:
total = 0;
% initial value is zero, in every while loop it will be incremented
while(input.hasNextInt()){
total += input.nextInt( );
}
Answer:



So then the value of the maximum I/O wait that can be tolerated is 0.720 or 72 %
Explanation:
Previous concepts
Input/output operations per second (IOPS, pronounced eye-ops) "is an input/output performance measurement used to characterize computer storage devices like hard disk drives (HDD)"
Solution to the problem
For this case since we have 4GB, but 512 MB are destinated to the operating system, we can begin finding the available RAM like this:
Available = 4096 MB - 512 MB = 3584 MB
Now we can find the maximum simultaneous process than can use with this:

And then we can find the maximum wait I/O that can be tolerated with the following formula:

The expeonent for p = 14 since we got 14 simultaneous processes, and the rate for this case would be 99% or 0.99, if we solve for p we got:



So then the value of the maximum I/O wait that can be tolerated is 0.720 or 72 %
Answer:
Python Code:
def validate_url(url):
#Creating the list of valid protocols and file name extensions
valid_protocols = ['http', 'https', 'ftp']
valid_fileinfo = ['.html', '.csv', '.docx']
#splitting the url into two parts
url_split = url.split('://')
isProtocolValid = False
isFileValid = False
#iterating over the valid protocols and file names for validity
for x in valid_protocols:
if x in url_split[0]:
isProtocolValid = True
break
for x in valid_fileinfo:
if x in url_split[1]:
isFileValid = True
break
#Returning the result if the URL has both valid protocol and file extension
return (isProtocolValid and isFileValid)
url = input("Enter an URL: ")
print(validate_url(url))
Explanation:
The image of the output code is attached. Hope it helps.