Answer:
Answered below
Explanation:
Scanner n = new Scanner(System.in);
System.out.print("Enter number of guests: ");
int numOfGuests = n.nextline();
double pricePerGuest = 35.0;
double totalAmount = numOfGuests * pricePerGuest;
String event = " ";
if ( numOfGuests >= 50){
event = "Large event";
}else{
event = "small event";
}
System.out.print(numOfGuests);
System.out.println(pricePerGuest);
System.out.println( totalAmount);
System.out.println(event);
Answer:
The page field is 8-bit wide, then the page size is 256 bytes.
Using the subdivision above, the first level page table points to 1024 2nd level page tables, each pointing to 256 3rd page tables, each containing 64 pages. The program's address space consists of 1024 pages, thus we need we need 16 third-level page tables. Therefore we need 16 entries in a 2nd level page table, and one entry in the first level page table. Therefore the size is: 1024 entries for the first table, 256 entries for the 2nd level page table, and 16 3rd level page table containing 64 entries each. Assuming 2 bytes per entry, the space required is 1024 * 2 + 256 * 2 (one second-level paget table) + 16 * 64 * 2 (16 third-level page tables) = 4608 bytes.
Answer: Logical security list
Explanation:Logical security list that is used for checking of the authentication and authorization for the organizations. It consist the safeguarding the software, security access of password and other keys, system auditing, stages of the security etc.
These factors helps in the maintaining the level of the security and only authorized access to place in the system of a particular organization.Other options are incorrect because physical security system consist of the guards , locks etc physical devices, response plane is the feedback strategy and whiltelist is related with the acceptable people.
Answer:
The answer is A4B₁₆ = 2635₁₀ = 101001001011₂
Explanation:
To convert from hexadecimal base system to binary base system, first you can do an intermediate conversion from hexadecimal to decimal using this formula:
where position of the x₁ is the rightmost digit of the number and the equivalents hexadecimal numbers to decimal:
- A = 10.
- B = 11.
- C = 12.
- D = 13.
- E = 14.
- F = 15.
A4B₁₆ = A*16²+4*16¹+B*16⁰ = 2560 + 64 + 11 = 2635₁₀
Now, you have the number transformed from hexadecimal to decimal. To convert the decimal number 2635 to binary: Divide the number repeatedly by 2, keeping track of each remainder, until we get a quotient that is equal to 0:
2635 ÷ 2 = 1317 + 1;
1317 ÷ 2 = 658 + 1;
658 ÷ 2 = 329 + 0;
329 ÷ 2 = 164 + 1;
164 ÷ 2 = 82 + 0;
82 ÷ 2 = 41 + 0;
41 ÷ 2 = 20 + 1;
20 ÷ 2 = 10 + 0;
10 ÷ 2 = 5 + 0;
5 ÷ 2 = 2 + 1;
2 ÷ 2 = 1 + 0;
1 ÷ 2 = 0 + 1;
Now, construct the integer part base 2 representation, by taking the remainders starting from the bottom of the list:
2635₁₀ = 101001001011₂
Answer:
umm... you did not discribe the anwsers, i cannot help you
