60 sec. = 1 min.
3,600 sec = 1 hr.
(60 sec x 60 minutes) 86,400 sec= 24 hr.
(3,600 x 24) Given : 16,000 (bytes per seconds, for mp3 )
86,400 X 16,000 = 1,382,400,000 bytes for 24 hrs.
1,382,400,000 = 1,382.4 mega bytes (Mb.) for 24 hrs of music
Looking for how many days are in 1,000,000 Mb.
1,000,000 / 1,382.4 = 723.379629 Days worth of music.
Answer:
public class Main
{
public static void main(String[] args) {
int userNum = 40;
while(userNum > 1){
userNum /= 2;
System.out.print(userNum + " ");
}
}
}
Explanation:
*The code is in Java.
Initialize the userNum
Create a while loop that iterates while userNum is greater than 1. Inside the loop, divide the userNum by 2 and set it as userNum (same as typing userNum = userNum / 2;). Print the userNum
Basically, this loop will iterate until userNum becomes 1. It will keep dividing the userNum by 2 and print this value.
For the values that are smaller than 1 or even for 1, the program outputs nothing (Since the value is not greater than 1, the loop will not be executed).
Answer:
A wand tool is to do that in an editing software.
Digital cellphones use Digital Compression to reduce the size of the channels that are required, by scattering the digital fragments of conversations over many channels
Answer:
Pseudocode is as follows:
// below is a function that takes two parameters:1. An array of items 2. An integer for weight W
// it returns an array of selected items which satisfy the given condition of sum <= max sum.
function findSubset( array items[], integer W)
{
initialize:
maxSum = 0;
ansArray = [];
// take each "item" from array to create all possible combinations of arrays by comparing with "W" and // "maxSum"
start the loop:
// include item in the ansArray[]
ansArray.push(item);
// remove the item from the items[]
items.pop(item);
ansArray.push(item1);
start the while loop(sum(ansArray[]) <= W):
// exclude the element already included and start including till
if (sum(ansArray[]) > maxSum)
// if true then include item in ansArray[]
ansArray.push(item);
// update the maxSum
maxSum = sum(ansArray[items]);
else
// move to next element
continue;
end the loop;
// again make the item[] same by pushing the popped element
items.push(item);
end the loop;
return the ansArray[]
}
Explanation:
You can find example to implement the algorithm.
