Software as a Service cloud model is ideal for Rita’s business. SaaS are office solutions that allow Rita’s small business to work more efficiently and in a more organized way. Most SaaS applications are used for invoicing and accounting, sales, performance monitoring, and overall planning. SaaS applications can save Rita money. They do not require the deployment of a large infrastructure at her location. As a result, it drastically reduces the upfront commitment of resources. Whoever manages SaaS’s IT infrastructure running the applications brings down fees for software and hardware maintenance. SaaS has generally been acknowledged to be safer than most on-premise software.
Answer:
for(String s:words)
if(s.endsWith("ing"))
System.out.println(s);
Explanation:
Create an enhanced for loop that iterates through the words array
Check if an element in words ends with "ing" using endsWith() method (Since it is said that strings are lowercase letters, we do not need to check it)
If you find one that ends with "ing", print the element
The correct answer is A. Familiar words for clues
Explanation:
Finding unfamiliar words is common while reading, especially in texts that belong to a specific field such as medicine, technology, etc. This can be handled through multiple strategies such as using a dictionary, guessing the meaning of the word based on its parts, and using context clues.
In this context, one of the easiest and most time-saving strategy is the use of context clues that implies using the familiar words as clues to guess the meaning of an unfamiliar word. This is effective because in most cases the meaning of an unknown word can be determined using the context of the word or words around the unknown word. Also, this strategy takes little time because you only need to analyze the sentence or paragraph where the unknown word is. Thus, the time-saving strategy to define unfamiliar words involves using familiar words for clues.
Answer:
#include <iostream>
#include <cstdlib>
using namespace std;
char grade(double marks){
if(marks>=90)
{
return 'A';
}
else if (marks >=80 && marks<90)
{
return 'B';
}
else if (marks >=70 && marks<80)
{
return 'C';
}
else if (marks >=60 && marks<70)
{
return 'D';
}
else if ( marks<60)
{
return 'F';
}
}
int main()
{
double marks;
cout <<"Ener marks";
cin >>marks;
char grd=grade(marks);
cout<<"Grae is "<<grd;
return 0;
}
Explanation:
Take input from user for grades in double type variable. Write function grade that takes a parameter of type double as input. Inside grade function write if statements defining ranges for the grades. Which if statement s true for given marks it returns grade value.
In main declare a variable grd and store function returned value in it.
