Answer:
This is because, the advertisers report the storage space in decimal while the computer reads the storage space in binary.
Explanation:
The advertisers report the storage space in decimal or base 10 because humans count in decimal, whereas the computer reports the storage space in binary or base 2.
Since the computer storage is in bytes and 8 bits equal 1 byte, It is easier to write a storage space of 1 kB as 1000 B but it is actually supposed to be 1024 B(2¹⁰). So, there is a discrepancy of 1024 B - 1000 B = 24 B.
As we go higher, the discrepancy increases. For example, 1 MB is advertised as 1000 kB = 1000000 B but is actually supposed to be 1024 kB = 1024 × 1024 B = 1048576 B. So, there is a discrepancy of 1048576 B - 1000000 B = 48576 B.
So, the actual number of bytes on the storage device is actually less than that reported due to the different number systems in which they are reported in. This discrepancy is less in memory cards or flash drives though in which the stated value of storage capacity might be the actual storage size.
Note that the base 10 or decimal system was chosen by advertiser since this is what consumers understand.
The annswer is word processing.
Answer:
see explaination
Explanation:
import java.util.ArrayList;
import java.util.Arrays;
public class Merge {
public static ArrayList<String> mergeList(ArrayList<String> lst1, ArrayList<String> lst2) {
ArrayList<String> result = new ArrayList<String>();
int i = 0, j = 0;
while (i < lst1.size() || j < lst2.size()) {
if (i < lst1.size() && (j >= lst2.size() || lst1.get(i).compareTo(lst2.get(j)) < 0)) {
result.add(lst1.get(i++));
} else {
result.add(lst2.get(j++));
}
}
return result;
}
public static void main(String[] args) {
ArrayList<String> lst1 = new ArrayList<>(Arrays.asList("Austin", "Dallas", "San Fransisco"));
ArrayList<String> lst2 = new ArrayList<>(Arrays.asList("Boston", "Chicago", "Denver", "Seattle"));
System.out.println(mergeList(lst1, lst2));
}
}
The operating system's memory protection routine intervenes and (usually) terminates the program if a program attempts to modify (or, sometimes, even to read) the contents of memory locations that do not belong to it.
Further Explanation
The memory protection routine is most commonly used in multi-programmed systems to prevent one process from affecting the availability of another. When a user opens up multiple processes, by default, they usually reside at the same time in the main memory. Sometimes, a program may attempt to access, modify, or read memory locations allocated to other processes. When this happens, the memory protection program jumps in. Keep in mind that the memory manager somehow works hand in hand with the memory protection routine. It protects the OS from being accessed by other processes and these processes from accessing one another. In addition, it helps save memory by allocating the same amount of memory to all running processes. The memory protection program, on the other hand, should be able to allow controlled sharing of memory among different processes and will usually terminate a program that tries to modify content of memory locations of that does not belong to it.
Learn More about Memory management
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Your answer is false :)
I hope this helps! :)
