Storing music files on a smartphone in order to maximize the number of
songs that can be stored is most appropriate for choosing lossy
compression over lossless compression.
<h3>What is Lossy compression?</h3>
Lossy compression refers to a type of data compression which results in
the file size being small. Most times the quality is reduced as a result of
this .
This gives rise to more space being made available for the storage of more
song files which is why it is most appropriate.
Read more about Lossy compression here brainly.com/question/18806025
Answer:
Un lenguaje de programación es un lenguaje formal (o artificial, es decir, un lenguaje con ... Todo esto, a través de un lenguaje que intenta estar relativamente próximo al lenguaje ... Artículo principal: Historia de los lenguajes de programación ... cada una representando lenguajes de programación surgidos en una época ...
Explanation:
Answer:
a simple IF statement using Regex
Explanation:
In any coding language, a good control procedure for this would be a simple IF statement using Regex. In the IF statement you can grab the account code and compare it to a regular expression that represents the correct format. IF the account code is in the correct format (matches the regular expression), then you go ahead and save the account code for use. Otherwise, you would output an error and ask for another account code. This will prevent the program from trying to use an account code that is not valid.
Answer:
Answered below
Explanation:
//Program is written using Java programming language.
Class Person {
private string firstName;
private string lastName;
void set firstName(string a){
firstName = a;
}
string getFirstName(){
return firstName;
}
void setLastname( string b){
lastName = b;
}
string getLastName( ){
return lastName;
}
void displayDetails( ) {
System.out.print(firstName);
System.out.print (lastName);
}
}
//Test program
Class Main{
public static void main(String args [] ){
Person person = new Person( )
person.setFirstName("Karen")
System.out.print(person.getFirstName)
person.displayDetails()
}
}
