Answer:
10G Ethernet
Explanation:
These are the options for the question;
A) 10BASE Ethernet
B) Gigabit Ethernet
C) Fast Ethernet
D) 10G Ethernet
From the question, we are informed about instance of my company deciding to upgrade the older office Ethernet network and needs the fastest speed possible but has decided against fiber optic cable. In this case my solution for this problem is getting
10G Ethernet. Ethernet can be regarded as traditional technology that connects devices in LAN(wired local area network) as well as WAN(wide area network) which allows them to have communication with each other through a protocol, this protocol is reffered to as common network language, it also be regarded as rules. 10 Gigabit Ethernet which is a technology ofgroup of computer networking that enables transmission of Ethernet frames at high rate of 10 gigabits per second. Therefore, 10G Ethernet is the solution since we need
the fastest possible speed.
The answer is a POST Diagnostic Card.
Although not a necessity, a POST card can help discover and report computer errors and conflicts that occur when you first turn on a computer and before the operating system. To be able to understand how a POST card works, one needs to be very familiar with the BIOS. The power-on self test (POST) is a series of various tests performed by a computer when you turn it on. If you have any issues that conflicts with the operating system and prevents the computer from booting, you can install a POST card in the available expansion slots. This card will monitor the entire boot process and report errors in coded numbers on a small LED panel on the card.
Answer:
import java.util.Scanner;
import java.util.Arrays;
import java.util.Random;
public class Main {
public static void main(String[] args) {
Scanner scan = new Scanner(System.in);
System.out.print("Enter low: ");
int low = scan.nextInt();
System.out.print("Enter high: ");
int high = scan.nextInt();
scan.close();
int rndnumbers[] = new int[10];
Random r = new Random();
for(int i=0; i<rndnumbers.length; i++) {
rndnumbers[i] = r.nextInt(high-low+1) + low;
}
for(int i=0; i<rndnumbers.length; i++) {
System.out.printf("%d: %d\n", i, rndnumbers[i]);
}
}
}
Answer:
1) The value of x will be 6.
2) The value of y will be 7.
Explanation:
1) The value of x will be 6.
The enum values are labeled by default from 1. This means that Sun = 1, Mon = 2, Tue = 3 and so on.
So, x = Mon = 2 and y = Sat = 6
x increases up to y = 6 in the while loop.
and then y also increments by 1.
2)So the value of y = 7.
You will need actual printf("Sun") or printf("Mon") for printing the actual text for the enum.
