Given a Fully Associative cache with 4 lines using LRU replacement. The word size is one byte, there is one word per block, and the main memory capacity is 16 bytes. Suppose the cache is initially empty and then the following addresses are accessed in the order listed: {3,4,3,7,7,12,4,8,3,12,7,9}.

Answer:

Call microsoft

Explanation:

Exel Is My Worst Nightmare Just Call Microsoft "I'd Hope They'd Have A Answer For That"

The IF function allows the user to make logical comparison among values.

The formula to enter in cell 15 is: [tex]\mathbf{=IF(\$A\$4 > \$A\$3, "yes", "no")}[/tex]

The syntax of an Excel IF function is:

[tex]\mathbf{=IF (logical\_test, [value\_if\_true], [value\_if\_false])}[/tex]

Where:

From the question, the cells to compare are:

So, the IF function that compares both cells is:

[tex]\mathbf{=IF(A4 > A3, "yes", "no")}[/tex]

The above formula checks if A4 is greater than A3.

The question requires that the cells are referenced using absolute cell referencing.

This means that, we make use of the dollar sign ($), when writing the cells.

So, the correct formula is:

[tex]\mathbf{=IF(\$A\$4 > \$A\$3, "yes", "no")}[/tex]

Read more about Excel formulas at:

https://brainly.com/question/1285762

Answer:

Replace/* Your code goes here */ with

for(int count = 0; count<4; count++)

{

if(keysList[count] > 40)

{

cout<<itemsList[count]<<" ";

}

}

Explanation;

The solution provides used an iteration to loop from the first toll the last.

The index of an array starts at 0

The index of the last element is calculated as total element - 1

Since the count of the array is 4, the last index is 4 - 1 = 3.

In the solution provides, a count integer variable is declared to iterate through the keysList array

Each element of keysList array is tested to know if it's greater than 40.

If yes the corresponding element in itemsList is printed followed by a space

Else

Nothing is done

The codes segment goes thus

for(int count = 0; count<4; count++)

{

if(keysList[count] > 40)

{

cout<<itemsList[count]<<" ";

}

}

Answer:

def test_prime(n):

   if (n==1):

       return False

   elif (n==2):

       return True;

   else:

       for x in range(2,n):

           if(n % x==0):

               return False

       return True              

print(test_prime(9))

Final answer:

The Python functions isPrime(number), HowManyPrimes(P), and HighestPrime(K) are used to check if a number is prime, count how many prime numbers are less than P, and find the highest prime number — respectively.

Explanation:

To determine if an integer number is prime, you can write a Python function isPrime(number). A simple way to do this is to check whether the number has any divisors other than 1 and itself. Here's a possible implementation:

def isPrime(number):
   if number <= 1:
       return False
   for i in range(2, int(number**0.5) + 1):
       if number % i == 0:
           return False
   return True

The function HowManyPrimes(P) can be written to count the number of prime numbers less than an integer P. Here is a function that uses isPrime to do that:

def HowManyPrimes(P):
   count = 0
   for number in range(2, P):
       if isPrime(number):
           count += 1
   return count

To find the highest prime number that is less than or equal to a given integer K, you can create the function HighestPrime(K):

def HighestPrime(K):
   for number in range(K, 1, -1):
       if isPrime(number):
           return number
   return None

Answer:

See Explaination

Explanation:

def merge(listA, listB):

if not listA:

return listB

if not listB:

return listA

# create a empty resulting list(merged list)

result = []

# check the first elements of list listA, listB)

if(listA[0]<listB[0]):

# append the first element of listA to result

result.append(listA[0])

# use recursion to get remaning elements

# listA is now reduced to listA[1:]

result.extend(merge(listA[1:],listB))

else:

# append the first element of listB to result

result.append(listB[0])

# use recursion to get remaning elements

# listB is now reduced to listB[1:]

result.extend(merge(listA,listB[1:]))

# return the resultant list

return result

def largest_sum(xs, x, y):

# get number of rows and columns

m = len(xs[0])

n = len(xs)

# if move is invalid

if(x>=n or y>=m):

return 0

# check if we have reached boundary then return xs[x][y]

if(x==(n-1) and y==(m-1)):

return xs[x][y]

else:

# return max of two possible moves(leftmove, down move)

# move left (in this increment x)

left = xs[x][y] + largest_sum(xs,x+1,y)

# move down (in this increment y)

down = xs[x][y] + largest_sum(xs,x,y+1)

# return the maximum of two possible moves

if(left>down):

return left

else:

return down

# test code

print("Output for sample test cases given in problem:")

print(merge([1,2,3], [4,5,6]))

print(largest_sum([[1,2],[3,0]],0,0))

print(largest_sum([[5,6,1],[2,3,3]],0,0))

print(largest_sum([[0,7,5],[6,-1,4],[-5,5,2]],0,0))

print(largest_sum([[0,7,5],[6,-1,4],[-5,5,2]],1,1))

Answer:

Explanation:

Firstly we declare a function that will take two inputs, message and student list (Line 1).

In the function, we first print the message and the original input student list (Line 2 - 3). Next, we use sort method to sort the input list and then output the sorted items from the list using a for loop (Line 5-10).

Next, we sort the list again by setting reverse = True and this will sort the list in descending order (Line 13). Again we use the similar way mentioned above to output the sorted items (in descending order) using a for loop (Line 14 -17)

We test the function using a sample student list (Line 18) and we shall get the output:

Welcome to new semester!

['Aaron', 'Bernice', 'Cody']

Our students are currently in alphabetical order

-Aaron-Bernice-Cody

Our students are currently in reverse alphabetical order

-Cody-Bernice-Aaron

Answer:

See explaination

Explanation:

Given that:

Consider the MIPS assembly language code segment given below.

I1: addi $s3, $s2, 5

I2: sub $s1, $s3, $s4

I3: add $s3, $s3, $s1

I4: Iw $s2, 0($s3)

I5: sub $s2, $s2, $s4

I6: sw $s2, 200($s3)

I7: add $s2, $s2, $s4

(a) Identify the data dependency in the following code:

RAW - Read after write: This dependency occurs when instruction I2 tries to read register before instruction I1 writes it.

WAW - Write after write: This dependency occurs when instruction I2 tries to write register before instruction I1 writes it

• Read after write (RAW)

I1: addi $s3, $s2, 5

I2: sub $s1, $s3, $s4

• Read after write (RAW)

I2: sub $s1, $s3, $s4

I3: add $s3, $s3, $s1

• Write after write (WAW)

I4: lw $s2, 0($s3)

I5: sub $s2, $s2, $s4

• Read after write (RAW)

I5: sub $s2, $s2, $s4

I6: sw $s2, 200($s3)

(b) Data hazards which can be solved by forwarding:

(1) Read after write (RAW)

I1: addi $s3, $s2, 5

I2: sub $s1, $s3, $s4

addi $s3, $s2,5 IF ID EXE MEM WB

sub $51, $s3, $s4 IF ID EXE MEM WB

(2) Read after write (RAW)

I2: sub $s1, $s3, $s4

I3: add $s3, $s3, $s1

sub $51, $s3. $54 IF | ID EXE |MEM WB

add $s3, $s3, $s1 IF | ID EXE | MEM | WB

(3) Read after write (RAW)

I5: sub $s2, $s2, $s4

I6: sw $s2, 200($s3)

sub $52, $s2, $54 IF ID |EXE MEM | WB

sw $52, 200($s3) IF ID EXE MEM WB

(c) Data hazards which can lead to pipeline stall:

• Write after write (WAW)

I4: lw $s2, 0($s3)

I5: sub $s2, $s2, $s4

lw $s2, 0($s3) IF ID EXE MEM WB

sub $s2, $s2, $s4 IF ID Stall EXE | MEM | WB

Answer:

See explaination

Explanation:

#include <iostream>

using namespace std;

/* Type your code for the function findWithinThreshold here. */

void findWithinThreshold(int argSource[], int argsSourseSize, int argThreshold, int argTarget[], int &argsTargetSize)

{

argsTargetSize = 0;

for (int i = 0; i < argsSourseSize; i++)

{

if (argSource[i] <= argThreshold)

argTarget[argsTargetSize++] = argSource[i];

}

}

/* Type your code for the function findWithinLimits here. */

void findWithinLimits(int argSource[], int argsSourseSize, int argLowLimit, int argHighLimit,int argTarget[],int &argTargetCount)

{

argTargetCount = 0;

for (int i = 0; i < argsSourseSize; i++)

{

if (argSource[i]>=argLowLimit && argSource[i] <= argHighLimit)

argTarget[argTargetCount++] = argSource[i];

}

}

int main() {

const int MAX_SIZE = 100;

int source[MAX_SIZE]; //integer array source can have MAX_SIZE elements inside it ;

// user may chose to use only a portion of it

// ask the user how many elements are going to be in source array and

// then run a loop to get that many elements and store inside the array source

int n;

cout << "How many elements: ";

cin >> n;

for (int i = 0; i < n; i++)

{

cout << "Enter " << (i + 1) << " element: ";

cin >> source[i];

}

int threshold, lower_limit, upper_limit;

/* Type your code to declare space for other required data like

target array (think how big it should be)

threshold

lower limit

upper limits

as well as the variable that will bring back the info regarding how many elements might be in target array

*/

int *target = new int[n];

int targetCount;

cout << "\nEnter thresold value: ";

cin >> threshold;

/* Type your code to get appropriate inputs from the user */

cout << "\nEnter lower limit: ";

cin >> lower_limit;

cout << "\nEnter upper limit: ";

cin >> upper_limit;

/* Type your code here to call/invoke the function findWithinThreshold here.. */

/* Type your code to print the info in target array - use a loop and think about how many elements will be there to print */

findWithinThreshold(source, n, threshold, target, targetCount);

cout << "\nElement in the threshold = " << targetCount << endl;

for (int i = 0; i < targetCount; i++)

cout << target[i] << "\n";

/* Type your code here to call/invoke the function findWithinLimits here.. */

/* Type your code to print the info in target array - use a loop and think about how many elements will be there to print */

findWithinLimits(source, n, lower_limit, upper_limit, target, targetCount);

cout << "\n\nElements with in the " << lower_limit << " and "<<upper_limit<<endl;

for (int i = 0; i < targetCount; i++)

cout << target[i] << " ";

cout << endl;

system("pause");

return 0;

}

The function findWithinThreshold identifies elements in an array less than a given threshold. It takes four parameters and outputs the target array and count of elements below the threshold. Sample C++ code is provided for implementation.

Function to Identify Elements within a Threshold :

To solve this problem, you need to write a function findWithinThreshold that identifies elements in an array which are less than a given threshold. The function takes four parameters:

Here's a sample implementation in C++:

In this function, argTargetCount is set to 0 initially, and for each element in argSource that is less than argThreshold, the element is added to argTarget and argTargetCount is incremented.

Answer:

A. The proposed solution delivers the required result and both optional desired results.

Explanation:

The IEEE 802.11a is the standard code of the wireless communication technology known as WIFI, it is wireless as it requires no cable to connect to a network device. The data transmission rate of the IEEE 802.11a is 1.5 - 54Mbps (mega bit per second). So, the employees in the TVBCA building can wireless connect to the network, which allows them to be mobile and they can also send large CAD files among themselves.

Answer:

import java.util.Scanner;

public class StrictlyIdentical

{

public static void main(String[] args) {

 Scanner input = new Scanner(System.in);

 int[] list1 = new int[5];

 int[] list2 = new int[5];

 System.out.println("Enter the numbers for list1: ");

 for (int i=0; i<5; i++) {

        list1[i] = input.nextInt();

    }

    System.out.println("Enter the numbers for list2: ");

    for (int i=0; i<5; i++) {

        list2[i] = input.nextInt();

    }

 if (equals(list1, list2))

     System.out.println("the lists are strictly identical.");

 else

     System.out.println("the two lists are not strictly identical.");

}

public static boolean equals(int[] array1, int[] array2) {

    boolean isIdentical = true;

    for (int i=0; i<5; i++) {

        if (array1[i] != array2[i])

            isIdentical = false;

    }

    return isIdentical;

}

}

Explanation:

Create a function called equals that takes two parameters, array1, and array2

Initialize the isIdentical as true, this will be our control variable to change its value if two arrays are not identical

Create a for loop that iterates through the arrays. If corresponding elements of the arrays are not equal, set isIdentical as false.

When the loop is done, return the isIdentical

Inside the main:

Declare two arrays

Ask the user to enter numbers for the arrays using for loop

Check if two arrays are identical using the equal function. Print the appropriate message

Answer:

a. see explaination

b. 0.632

Explanation:

Packet switching is a method of grouping data that is transmitted over a digital network into packets. Packets are made of a header and a payload.

See attachment for the step by step solution of the given problem.

This response translates a C++ function to MIPS assembly language, detailing each MIPS instruction with comments and explanations. The step-by-step process converts high-level constructs like loops and conditions into MIPS equivalents. Detailed MIPS code is provided and explained for better understanding.

Translation of C++ Program to MIPS Assembly

The provided C++ function performs an operation on an array, and we will convert each instruction to its MIPS equivalent.

C++ Code:

#include  using namespace std;

void myfunction(int arr[], int n)

{ int writes = 0;

for(int start = 0; start <= n-2; start++)

{ int item = arr[start];

int pos = start;

for(int i = start+1; i < n; i++)

if (arr[i] < item) pos++;

// ... for swapping and updating ... }}

MIPS Assembly Code:

# Comment: function myfunction(int arr[], int n) defined start:

# Commentary on each variable:

# arr = $a0 (base address of the array)

# n = $a1 (number of elements in array)

# writes = $s0

# start = $s1

# item = $s2

# pos = $s3

# i = $s4

# Clear writes variable to 0 li $s0, 0

# Outer loop for start = 0 to n-2 outer_loop: move $s1, $zero

# initialize start move $s1, $t0

# copy start from $t0

# Check condition start <= n-2: sub $t2, $a1, 2 ble $s1, $t2, continue_outer_loop j end_of_program continue_outer_loop:

# int item = arr[start] sll $t1, $s1, 2

# start * 4 add $t3, $a0, $t1

# arr[start] lw $s2, 0($t3)

# item = arr[start]

# int pos = start move $s3, $s1

# pos = start

# Inner loop for(int i = start+1; i < n; i++) inner_loop: add $s4, $s1, 1

# i = start + 1 move $t4, $a1

# $t4 = n ble $s4, $t4, continue_inner_loop j update_outer_loop continue_inner_loop:

# if (arr[i] < item) lw $t5, 0($t3)

# arr[i] blt $t5, $s2, increment_pos j end_inner_loop increment_pos: add $s3, $s3, 1

# pos++ end_inner_loop: j inner_loop update_outer_loop:

# ... code for swapping and updating ... j outer_loop end_of_program:

Here are the detailed explanations for each MIPS instruction:

li $s0, 0: Initialize counter 'writes' to 0.

move $s1, $zero: Initialize 'start' to 0.

sll $t1, $s1, 2: Calculate memory offset for arr[start] (start*4).

add $t3, $a0, $t1: Get the address of arr[start].

lw $s2, 0($t3): Load value of arr[start] into 'item'.

move $s3, $s1: Initialize 'pos' with 'start'.

add $s4, $s1, 1: Set 'i' to start+1.

blt $t5, $s2,: Compare arr[i] with 'item'.

add $s3, $s3, 1: Increment 'pos' if arr[i] is less than 'item'.

Answer:

number = int(input("Enter a number: "))

product = 1

while number > 0:

   digit = number % 10

   number -= digit

   number /= 10

   product *= digit

print(int(product))

Explanation:

*The code is in Python

Ask the user for an integer

Initialize the product variable that will hold the multiplication of the digits

Initialize a while loop that iterates until number is greater than 0

To find the digit of a number, use number modulo 10. Then subtract the result from the number and divide the new number by 10.

Multiply each digit and hold the value in the product

When the loop is done, print the product

Answer:

The complete work sheet is attached, also use these formulas below for more guidance:

E13 =IF(C13="No",$F$2,$F$4) copy down to E18

F13 =VLOOKUP(B13,$A$7:$B$10,2,0)*$B$5 copy down to F18

H13 =SUM(D13:G13) copy down to H18

I2 =AVERAGE(H13:H18)

I3 =MIN(H13:H18)

I4 =MAX(H13:H18)

Answer:

def select_short_strings(string_list):

   new_list = []

   for s in string_list:

       if len(s) < 20:

           new_list.append(s)

   return new_list

lst = ["apple", "I am learning Python and it is fun!", "I love programming, it is easy", "orange"]

print(select_short_strings(lst))

Explanation:

- Create a function called select_short_strings that takes one argument string_list

Inside the function:

- Initialize an empty list to hold the strings that are less than 20

- Inside the loop, check the strings inside string_list has a length that is smaller than 20. If found one, put it to the new_list.

- When the loop is done, return the new_list

- Create a list to check and call the function

Answer:

# the solution function is defined

# it takes a list as parameter

def solution(string_list):

# an empty new list is initialized

new_list = []

# a loop that loop through the splitted_list

# it checks if the length of each string is less than 20

# if it is less than 20

# it is attached to the new list

for each_string in string_list:

if(len(each_string) < 20):

new_list.append(each_string)

# the new list is printed

print(new_list)

# the function is called with a sample list

solution(["The", "player", "determined", "never", "compromised"])

Explanation:

The program is written in Python and it is well commented.

The following code will be used to calculate the given requirement.

Explanation:

This function will return string containing an HTML unordered list with each of in_list strings as a list item

def make_html_unordered_list(in_list):

   result = '<ul>\n'

   for item in in_list:

       result += '<li>' + item + '</li>\n'

   return result + '</ul>'

# Testing the function here. ignore/remove the code below if not required

print(make_html_unordered_list(['CS 105', 'CS 125', 'IS 206']))

Note: The only whitespace that is important is those contained in the given strings.

The program is an illustration of loops.

Loops are used to perform repetitive and iterative operations.

The function in Python where comments are used to explain each line s as follows:

#This defines the function

def make_html_unordered_list(in_list):

   #This creates the opening tag of an HTML unordered list

   result = '<ul>\n'

   #This iterates through the list

   for item in in_list:

       #This creates the unordered list, without the last closing tag

       result += '<li>' + item + '</li>\n'

   #This returns the unordered list, with the last closing tag

  return result + '</ul>'

Read more about similar programs at:

https://brainly.com/question/14282285

To return the integer at a given index from a comma-separated string, split the string into a list and then convert the desired element to an integer.

To obtain the indexth value from a comma-separated string of integers, you can use the split() method to create a list of substrings, then use the int() function to convert the desired element to an integer. Here's an example of how you could write such a function:

Answer:

See the attached file for the answers

Explanation:

Find attached for the explanation

Amnswer:

Exposure of kids to unwanted pages, poor power system and refusal to backup works are some of the risk analysts

Explanation:

RISK ANALYSIS

1. Kids access to unwanted pages and sites if online are a big risk to private information and exposure to threat from hackers and information scammers

2. Poor power system results in an abrupt shutdown of system which is not encouraged. This is commonly responsible for system breakdown and other malfunctioning of the system

3. Lack of backup structure when not put in place are a very good avenue to home network risk which could have a very harmful effect to the home network activities.

DISASTER RECOVERY PLANS

1. Establish a system recovery and restore plans that could help you roll back actions when there is a system error that prevents the network from booting or starting up.

2. Establish a system restriction plan on your home network when kids and children are involved in it usage else vital information and configurations could be hampered. Focus in this context should also be geared towards website control system and constant system monitoring structure.

3. Power backup plans must include setting up a steady power source which should serve as alternatives to inconsistent power outage, irregular power supply and possible related issues. Such power structure should be able to mitigate for any network activities that requires a long time to backup

4. Other home network activities that requires consistent and constant backup should be done through external storage devices and through on line drives in order to ensure that document are not lost when emergency breakdown occurs.

SYSTEM ANTIVIRUS AND ANTI SPYWARES

From research some of the most used antivirus with spyware's features that has been largely used both online band offline include;

1. The Norton 360

2. Norton security

3. Avast

4. Avira ad lots more.

Answer:

MyString= input('Enter string: ')

string_capitalize(MyString)

print(MyString)

def string_capitalize(str):

        str= str.capitalize()

        return str

Explanation:

This python code is learnt to access string value from users and capitalize all characters in the beginning of each sentence.

Answer:

ESS

Explanation:

Employee self-service (ESS) is a widely used human resources technology that enables employees to perform any job-related functions, such as applying for reimbursement, updating personal information and accessing company benefits information -- which was once largely paper-based, or otherwise would have been maintained by management or administrative staff.

Employee self-service is often available through the employer's intranet or portal. It can also be part of larger human capital management (HCM), enterprise resource planning (ERP) or benefits administration software, which is often delivered via SaaS platforms. Employee self-service software, once sold as a stand-alone product, is now usually incorporated into more comprehensive HR tech systems.

Employee self-service systems are being optimized for mobile more and more on social media-like platforms, and are often part of larger employee engagement strategies, which can include wellness programs, recognition, learning management systems and organization-wide social activities.

A student seeks help with assembly programming to create a procedure generating random strings of capital letters. The procedure must handle varying lengths and return a pointer to the generated string. The driver program calls this procedure multiple times to output different strings.

The student's question relates to the field of assembly programming and specifically deals with creating a procedure that generates a random string. The procedure CreateRandomString receives the length of the string in the EAX register and returns a pointer to the string in the ESI register. The driver program calls this procedure 20 times to generate random strings with lengths varying between 1 and 100 characters, using uppercase ASCII character codes (65-90).To implement this in assembly, pseudocode for the CreateRandomString procedure would set up the loop to iterate L times, each time generating a random number within the range of uppercase letter ASCII codes and storing it in the resulting array. After the string is crafted, the pointer to this array is placed in the ESI register for the calling program to use. The calling or driver program uses a loop to generate 20 such strings, each time generating a new length at random from 1 to 100, then calling CreateRandomString and outputting the result to the console.

Answer:

1. Repetition control flow

3. Accessing the elements of a row in a 2D array that is heap allocated as an array of arrays

Explanation:

1. Elements in 1D memory are stored in contiguous memory location thus it follows spatial locality. ( RIGHT OPTION)

2. Binary search in an array does not follow spatial locality because the elements in an array with the binary search are not accessed in a contiguous manner but in a non-contiguous manner. HENCE IT IS NOT RIGHT OPTION

3. Sequencing control also follows spatial locality ( RIGHT OPTION)

Answer:

Step 1 : For TFF with asynchronous reset, the verilog code is :

'timescale 1ns/100ps

module tff1( input t,clk,reset, output reg q );

if (reset ) begin

always at (posedge clk ) begin

q <= #4 not q;

end

end else begin

q <= 1'b0;

end

end module

Step 2 : For TFF with synchronous reset, just include reset condition inside always statement as shown :

always at(posedge clk ) begin

if ( reset ) then

q <= #4 not q;

end else begin

q <= 1,b0'

end

end

Step 3 : For developing a DFF from a TFF , you need to have a feedback loop from ouput to input . Make sure you assign the wires correctly including the signal direction . Combine both the input D and ouptut of TFF using XOR and input it to the T.

module dff (input d, clk , reset ,output reg q )

wire q;

reg t;

tff1 ( t, clk, reset , q ); //module instantiation

xor ( t,q,d);

end module

For synchronous reset , you can just replace the tff asynchronous module with synchronous module

Step 4 : For obtaining JK FF using the DFF , we just to config the 4x1 MUX such that the required output is generated

module JKFF ( input j,k ,clk, reset , output reg q)

DFF ( d ,clk, reset ,q )

reg d;

case (j,k)

when "00" then d <= q;

when "01" then d <= 1'b0;

when "10" then d <= 1'b1;

when "11" then d <= #4 not d;

default : d <= 1'bX;

end module;

Final answer:

To encrypt the plaintext message, Bob uses Alice's public key and a random element. The ciphertext is obtained by multiplying the plaintext message with the public key. To decrypt the ciphertext, Alice uses her private key by multiplying the ciphertext with the private key polynomial and reducing modulo p.

Explanation:

To encrypt the plaintext message, Bob needs to use Alice's public key along with a random element. The ciphertext is obtained by multiplying the plaintext message with the public key. In this case, the plaintext message m(x)= 1+ X − X2 − X3 − X6 and the random element r(x) = −1 + X2 − X5 + X6. By multiplying these polynomials, Bob obtains the ciphertext c(x)= -1 - X - X^2 + 4X^3 + 8X^4 - 5X^5 - 30X^6 + 5X^7 - 6X^9 - 3X^10 + 8X^11 + 13X^12 + 2X^13 - 6X^14 + 26X^15 - 16X^16.

To decrypt the ciphertext, Alice uses her private key. She multiplies the ciphertext with her private key polynomial and reduces modulo p, resulting in the decrypted message. The private key polynomial for Alice is f(x) = −1 + X − X2 + X4 + X6. By multiplying the ciphertext c(x) with the private key polynomial f(x) modulo 3, the decrypted message is obtained: d(x)= 1+ X − X2 − X3 − X6.

Answer:

In daily routine I use different physical storage media such as Hard disk drive, USB (Flash Memory). Data can be accesses through these devices at the speed of up to 100 Mbps and 1 Mbps respectively.

Explanation:

There are different storage media that we can use to transfer data between different computer devices or store data permanently such as hard disk drive, floppy disk, optical storage devices (CD and DVD), USB (Flash Memory) and SD Cards. These storage medias have different purpose and different applications of use. All these devices works on different accessing speed of data.

In my daily routine, I usually use Hard disk drive for permanent storage of my important data and USB device for transferring data between different device or to keep data temporarily. Hard disk drives are available in terabyte size now a days while USB drives are available in the range 32 to 64 GB. Hard disk drive has data accessing speed of up to 100 Mbps and USB drive has almost speed of up to 1 Mbps to access data.

Answer:

Number of fragmented datagrams: 5

Explanation:

Since the MTU is 700 bytes,the payload will be 680 bytes and Up header will be 20 bytes.

The datagram is of size 3000 bytes. Therefore this datagram is fragmented to send it in 700 bytes MTU.

The first four fragments has the size 680 bytes of payload(4*680= 2720 bytes). The remaining 280 bytes will be sent in the 5th fragment.

The offset value gives how much was sent before the current fragment in terms of 8-byte blocks.

For first fragment it will be 0, as no data was sent before this first fragment.

For second fragment, 680 bytes(680/8=85 8-bytes block) were sent in first fragment before this current fragment. Hence the value will be 0+85=85.

For third fragment it is : 0+85+85=170

Number of fragmented datagrams: 5

Answer:

Following are program to this question:

#include <stdio.h> //defining hader file  

struct PatientData //defining structure PatientData

{

  int heightInches, weightPounds; //defining integer variable

};

int main() //defining main method

{

struct PatientData pd= {63,115}; //creating structure object and assign value

printf("%d %d", pd.heightInches, pd.weightPounds); //print value

return 0;

}

Output:

63 115

Explanation:

The program to this question can be described as follows:

Complete Data:

Job No.          Devices Allocated       Maximum Required       Remaining Need

Job 1                          6                                        8                                    X

Job 2                          2                                        9                                    X

Job 3                          5                                        8                                    X

Job 4                          3                                         5                                   X    

Answer:

a) Number of available devices = 2

b) Remaining need for job 1 = 2

Remaining need for job 2 = 7

Remaining need for job 3 = 3

Remaining need for job 4 = 2

c) The system is safe

Explanation:

a) Calculate the number of available devices

Number of available devices = (Maximum number of devices) - (Total number of devices allocated)

Maximum number of devices  = 18

Total number of allocated devices = (6 + 2 + 5 + 3)

Total number of allocated devices = 16

Number of available devices = 18 - 16

Number of available devices = 2

b) Determine the remaining needs for each job in each system

Remaining needs = (Maximum Required - Number of devices allocated)

For Job 1

Remaining need = 8 - 6 = 2

For Job 2

Remaining need = 9 - 2 = 7

For Job 3

Remaining need = 8 - 5 = 3

For Job 4

Remaining need = 5 - 3 = 2

c) Determine whether the system is safe or unsafe

For the system to be safe, Need ≤ Available

Where New Available = Available + Allocation

For Job 1

Need = 2

Available devices = 2

2 ≤ 2   (Job 1 is safe)

New available = 2 + 6 = 8

For Job 2

Need = 7

Available devices = 8

7 ≤ 8   (Job 2 is safe)

New available = 2 + 8 = 10

For Job 3

Need = 3

Available devices = 10

3 ≤ 10   (Job 3 is safe)

New available = 5 + 10 = 15

For Job 4

Need = 2

Available devices = 15

2 ≤ 15   (Job 2 is safe)

New available = 3 + 15 = 18

Since all the jobs are safe, the system is safe

Answer:

See explaination

Explanation:

#include<stdio.h>

#include<string.h>

#include<stdlib.h>

int main()

{

FILE *fp;

char section[10][2], score[10][10],buf[10][4], buf1[4];

int i=0, j = 0, k = 0, l = 0, c1 = 0, c2 = 0, c3 = 0, sec;

float avg1 = 0 ,avg2 = 0, avg3 = 0 ,avg[3] ,max;

fp = fopen("scores.dat","r");

/* read data from scores.dat file and store into buf */

while(fscanf(fp, "%s", buf[i++])!=EOF);

/* separate scores and sections */

for(j=1; j<i; j++){

if(j%2!=0){

strcpy(section[k++], buf[j-1]);

}

else{

strcpy(score[l++], buf[j-1]);

}

}

/* calculate avgerage */

for(i=0;i<7;i++){

if(strcmp(section[i], "1")){

avg1 = avg1 + atof(score[i]);

c1++;

}

if(strcmp(section[i], "2")){

avg2 = avg2 + atof(score[i]);

c2++;

}

if(strcmp(section[i], "1")){

avg3 = avg3 + atof(score[i]);

c3++;

}

}

avg[0] = avg1/c1;

avg[1] = avg2/c2;

avg[2] = avg3/c3;

max = avg[0];

for(i=0;i<3; i++)

if(avg[i] > max){

max = avg[i];

sec = i;

}

printf("Sectrion 1 Avg: %f\nSection 2 Avg: %f\nSection 3 Avg: %f\nHighest Avg by:Section %s\n",avg[0],avg[1],avg[2],section[sec]);

}

Answer:

#include<iostream>

#include<conio.h>

using namespace std;

main()

{

int n;

cout<<"enter the size of array"<<endl;

cin>>n;

int a[n],count=0;

for (int i=0;i<n;i++)

{

 cout<<"\nenter the values in array"<<i<<"=";

 cin>>a[i];  

}

for (int j=0;j<n;j++)

{

 if (a[j]>=100)

 {

  count=count+1;

 }

}

cout<<"Total Values greater than 100 are ="<<count;

getch();

}

Explanation:

In this program, an array named as 'a' has been taken as integer data type. The array haves variable size, user can change the size of array as per need or demand.

Values will be inserted by user during run time. This program will check that how many values are greater than and equal to 100 in array. The number of times 100 or greater value will appear in array, it count the value in variable "count". After complete traversal of array the result will shown on output in terms of total number of count of value greater than or equal to 100.

Answer:

Explanation:

A catalog is a directory where information about the location of file or database is stored

(a) It is good for DBMS to upgrade the catalog automatically when there is a change in the database structure.

It has the following benefits;

- The updated form of the database can be made available and accessible to other users

- The data of the database will remain in a stable and consistent state

- There will not be a need to execute the update query to update the database.

-If there is no update automatically, only users that made changes will be able to have access to the database

(b) Yes, users can cause problems by up dating the catalog manually.

The following will be the disadvantage when the catalog is updated by users;

- Updated database will not be seen until the user update manually by themselves

- Persistence and consistency of problems

Answer:

The catalog is defined as a directory in which the information about the database is stored. This catalog contains information about the location of any file or database. That DBMS update its catalog automatically when any change in the database is made is a good choice, since any user could enter and consult either a file or a database that is kept updated. If DBMS does not automatically update the database, only users who make any modifications to that database will be able to access that updated information, this would be a typical inconvenience if users update the catalog manually.

Explanation: