The answer is DS-30A. The Roland DS30A is a 24-bit Digital Reference Monitor which can add pristine 24-bit/96kHz. They are small speakers.
Answer:
the answer for the question is D.
Answer:
It we were asked to develop a new data compression tool, it is recommended to use Huffman coding since it is easy to implement and it is widely used.
Explanation:
The pros and the cons of Huffman coding
Huffman coding is one of the most simple compressing encoding schemes and can be implemented easily and efficiently. It also has the advantage of not being patented like other methods (e.g. arithmetic codingfor example) which however are superior to Huffman coding in terms of resulting code length.
One thing not mentioned so far shall not be kept secret however: to decode our 96 bit of “brief wit” the potential receiver of the bit sequence does need the codes for all letters! In fact he doesn’t even know which letters are encoded at all! Adding this information, which is also called the “Huffman table” might use up more space than the original uncompressed sentence!
However: for longer texts the savings outweigh the added Huffman table length. One can also agree on a Huffman table to use that isn’t optimized for the exact text to be transmitted but is good in general. In the English language for example the letters “e” and “t” occur most often while “q” and “z” make up the least part of an average text and one can agree on one Huffman table to use that on average produces a good (=short) result. Once agreed upon it doesn’t have to be transmitted with every encoded text again.
One last thing to remember is that Huffman coding is not restricted to letters and text: it can be used for just any symbols, numbers or “abstract things” that can be assigned a bit sequence to. As such Huffman coding plays an important role in other compression algorithms like JPG compression for photos and MP3 for audio files.
The pros and the cons of Lempel-Ziv-Welch
The size of files usually increases to a great extent when it includes lots of repetitive data or monochrome images. LZW compression is the best technique for reducing the size of files containing more repetitive data. LZW compression is fast and simple to apply. Since this is a lossless compression technique, none of the contents in the file are lost during or after compression. The decompression algorithm always follows the compression algorithm. LZW algorithm is efficient because it does not need to pass the string table to the decompression code. The table can be recreated as it was during compression, using the input stream as data. This avoids insertion of large string translation table with the compression data.
Answer:
A stores code for later re-use
Explanation:
To find - What is the purpose of saving code snippets?
Solution -
Code Snippet" is used to describe a small portion of re-usable source code, machine code, or text.
They allow a programmer to avoid typing repetitive code during the course of routine programming.
So,
The correct option is - A stores code for later re-use
Answer:
// code in C++
#include <bits/stdc++.h>
using namespace std;
// main function
int main()
{
// variables
int sum_even=0,sum_odd=0,eve_count=0,odd_count=0;
int largest=INT_MIN;
int smallest=INT_MAX;
int n;
cout<<"Enter 10 Integers:";
// read 10 Integers
for(int a=0;a<10;a++)
{
cin>>n;
// find largest
if(n>largest)
largest=n;
// find smallest
if(n<smallest)
smallest=n;
// if input is even
if(n%2==0)
{
// sum of even
sum_even+=n;
// even count
eve_count++;
}
else
{
// sum of odd
sum_odd+=n;
// odd count
odd_count++;
}
}
// print sum of even
cout<<"Sum of all even numbers is: "<<sum_even<<endl;
// print sum of odd
cout<<"Sum of all odd numbers is: "<<sum_odd<<endl;
// print largest
cout<<"largest Integer is: "<<largest<<endl;
// print smallest
cout<<"smallest Integer is: "<<smallest<<endl;
// print even count
cout<<"count of even number is: "<<eve_count<<endl;
// print odd cout
cout<<"count of odd number is: "<<odd_count<<endl;
return 0;
}
Explanation:
Read an integer from user.If the input is greater that largest then update the largest.If the input is smaller than smallest then update the smallest.Then check if input is even then add it to sum_even and increment the eve_count.If the input is odd then add it to sum_odd and increment the odd_count.Repeat this for 10 inputs. Then print sum of all even inputs, sum of all odd inputs, largest among all, smallest among all, count of even inputs and count of odd inputs.
Output:
Enter 10 Integers:1 3 4 2 10 11 12 44 5 20
Sum of all even numbers is: 92
Sum of all odd numbers is: 20
largest Integer is: 44
smallest Integer is: 1
count of even number is: 6
count of odd number is: 4
