Answer:
You need to first use the Sum function to add up all the costs of September.
Then divide each September cost by the grand total that you got. After that format the last column to be percentages.
Look at the attached file for the formulas used.
Answer: Result:Enter the number of quarters in the jar: 4
Enter the number of dimes in the jar: 5
Enter the number of nickels in the jar: 2
Enter the number of pennies in the jar: 5
Total value is 1 dollars and 65 cents
Explanation: import java.util.Scanner;public class Coins { public static void main(String[]args) { int quarters, dimes, nickels, pennies; int total;Scanner scan = new Scanner (System.in);System.out.print("Enter the number of quarters in the jar: "); quarters = scan.nextInt();System.out.print("Enter the number of dimes in the jar: ");dimes = scan.nextInt(); System.out.print("Enter the number of nickels in the jar: ");nickels= scan.nextInt(); System.out.print("Enter the number of pennies in the jar: ");pennies = scan.nextInt(); int total_cents = 25*quarters + dimes*10 + nickels*5 + pennies;total =total_cents/100;total_cents = total_cents %100;System.out.println("Total value is " + total + " dollars and " + total_cents + " cents ");}}
Result:Enter the number of quarters in the jar: 4
Enter the number of dimes in the jar: 5
Enter the number of nickels in the jar: 2
Enter the number of pennies in the jar: 5
Total value is 1 dollars and 65 cents
The user can use the Play store on there android device
Answer:
MAN IN THE MIDDLE
Explanation:
Man in the middle attack is an attack where the attacker secretly relays and possibly alters the communications between two parties. When data is sent between a computer and a server, a cybercriminal can get in between and spy or eavesdrop.
A man in the middle attack can positions himself in a conversation between a user and an application in other to impersonate one of the parties, making it appear as if a normal exchange of information is underway.
In a man-in-the-middle cyber-attack, the attacker places themselves in between two devices and monitor packets from the network, modifies them, and inserts them back into the network without the other parties knowing.
Answer:
Given that:
A= 40n^2
B = 2n^3
By given scenario:
40n^2=2n^3
dividing both sides by 2
20n^2=n^3
dividing both sides by n^2 we get
20 = n
Now putting n=20 in algorithms A and B:
A=40n^2
= 40 (20)^2
= 40 * (400)
A= 16000
B= 2n^3
= 2 (20)^3
= 2(8000)
B= 16000
Now as A and B got same on n = 20, then as given:
n0 <20 for n =20
Let us take n0 = 19, it will prove A is better than B.
We can also match the respective graphs of algorithms of A and B to see which one leads and which one lags, before they cross at n= 20.
