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:
It will be a java code.
Explanation:
import java.util.Scanner;
public class StringInputStream {
public static void main (String [] args) {
Scanner inSS = null;
String userInput = "Jan 12 1992";
inSS = new Scanner(userInput);`
String userMonth = "";
int userDate = 0;
int userYear = 0;
/* Your solution goes here */
System.out.println("Month: " + userMonth);
System.out.println("Date: " + userDate);
System.out.println("Year: " + userYear);
return;
}
}
Answer:
- public class FindDuplicate{
-
- public static void main(String[] args) {
- Scanner input = new Scanner(System.in);
-
- int n = 5;
- int arr[] = new int[n];
-
- for(int i=0; i < arr.length; i++){
- int inputNum = input.nextInt();
- if(inputNum >=1 && inputNum <=n) {
- arr[i] = inputNum;
- }
- }
-
- for(int j =0; j < arr.length; j++){
- for(int k = 0; k < arr.length; k++){
- if(j == k){
- continue;
- }else{
- if(arr[j] == arr[k]){
- System.out.println("True");
- return;
- }
- }
- }
- }
- System.out.println("False");
- }
- }
Explanation:
Firstly, create a Scanner object to get user input (Line 4).
Next, create an array with n-size (Line 7) and then create a for-loop to get user repeatedly enter an integer and assign the input value to the array (Line 9 - 14).
Next, create a double layer for-loop to check the each element in the array against the other elements to see if there is any duplication detected and display "True" (Line 21 - 22). If duplication is found the program will display True and terminate the whole program using return (Line 23). The condition set in Line 18 is to ensure the comparison is not between the same element.
If all the elements in the array are unique the if block (Line 21 - 23) won't run and it will proceed to Line 28 to display message "False".
Answer:
C++.
Explanation:
<em>Code snippet.</em>
#include <map>
#include <iterator>
cin<<N;
cout<<endl;
/////////////////////////////////////////////////
map<string, string> contacts;
string name, number;
for (int i = 0; i < N; i++) {
cin<<name;
cin<<number;
cout<<endl;
contacts.insert(pair<string, string> (name, number));
}
/////////////////////////////////////////////////////////////////////
map<string, string>::iterator it = contacts.begin();
while (it != contacts.end()) {
name= it->first;
number = it->second;
cout<<word<<" : "<< count<<endl;
it++;
}
/////////////////////////////////////////////////////////////////////////////////////////////////////////
I have used a C++ data structure or collection called Maps for the solution to the question.
Maps is part of STL in C++. It stores key value pairs as an element. And is perfect for the task at hand.
