What is resource Management in Wireless Communication ? Explain its Advantages?

def reverse_list (num_list):

try:

if num_list.isdigit() == True:

return num_list[::-1]

elif num_list < 0:

return "Negative numbers aren't allowed!"

except ValueError:

return "Invalid input!"

user_list = list (int (input ("Please enter ten random positive numbers: " )))

print (reverse_list (user_list))

Answer:

Basic characteristics of OOAD and compression with SAD can be defined as follows:

Explanation:

The Object-oriented analysis is described as a design and analysis system like an application, that uses Oops programming. It is the measurement model, that is the presentation, which became a part of the object-based review.  

The features of OOAD can be defined as follows.  

The difference in OOAD and SAD can be defined as follows:

Answer:

we have that it grows more quickly than linear.

Explanation:

It will still work if they are divided into groups of 77, because we will still know that the median of medians is less than at least 44 elements from half of the \lceil n / 7 \rceil⌈n/7⌉ groups, so, it is greater than roughly 4n / 144n/14 of the elements.

Similarly, it is less than roughly 4n / 144n/14 of the elements. So, we are never calling it recursively on more than 10n / 1410n/14 elements. T(n) \le T(n / 7) + T(10n / 14) + O(n)T(n)≤T(n/7)+T(10n/14)+O(n). So, we can show by substitution this is linear.

We guess T(n) < cnT(n)<cn for n < kn<k. Then, for m \ge km≥k,

\begin{aligned} T(m) & \le T(m / 7) + T(10m / 14) + O(m) \\ & \le cm(1 / 7 + 10 / 14) + O(m), \end{aligned}

T(m)

​

≤T(m/7)+T(10m/14)+O(m)

≤cm(1/7+10/14)+O(m),

​

therefore, as long as we have that the constant hidden in the big-Oh notation is less than c / 7c/7, we have the desired result.

Suppose now that we use groups of size 33 instead. So, For similar reasons, we have that the recurrence we are able to get is T(n) = T(\lceil n / 3 \rceil) + T(4n / 6) + O(n) \ge T(n / 3) + T(2n / 3) + O(n)T(n)=T(⌈n/3⌉)+T(4n/6)+O(n)≥T(n/3)+T(2n/3)+O(n) So, we will show it is \ge cn \lg n≥cnlgn.

\begin{aligned} T(m) & \ge c(m / 3)\lg (m / 3) + c(2m / 3) \lg (2m / 3) + O(m) \\ & \ge cm\lg m + O(m), \end{aligned}

T(m)

​

≥c(m/3)lg(m/3)+c(2m/3)lg(2m/3)+O(m)

≥cmlgm+O(m),

​

therefore, we have that it grows more quickly than linear.

Answer:

The work associated with software engineering can be categorized into three generic phases,regardless of application area, project size, or complexity namely the_____________ phase which focuses on what, the______________ phase which focuses on how and the_____________ phase which focuses on change ? i. support ii. development iii. definition

[A]. 1, 2, 3

[B]. 2, 1, 3

[C]. 3, 2, 1. ✅✅

[D]. 3, 1, 2

Answer:

I think the above information will help you a lot.

Have a nice day.

Please read the screen shot brainly is not letting me write the whole thing

Answer:

True

Explanation:

Please mark me as the brainiest

Answer:

HELLOOOO

Alr lets start with steps by dividing by 2 again and againn..

265 / 2 = 132 ( rem = 1 )

132 / 2 = 66 ( rem = 0 )

66/2 = 33 ( rem = 0 )

33/2 = 16 ( rem = 1 )

16/2 = 8 ( rem = 0 )

8/2 = 4 ( rem = 0 )

4/2 = 2 ( rem = 0 )

2/2 = 1 ( rem = 0 )

1/2 = 0 ( rem = 1 )

now write all the remainders from bottom to up

100001001

is ur ans :)))

Answer:

Round trip times required =  log2N

Explanation:

The round-trip times required before TCP can send N segments using a slow start is log2N. we can arrive at this by looking at the mode of operation of TCP which is  at the 1st time of using a TCP it starts the congestion window as 1 then it  sends an initial segment. When the acknowledgement of the initial segment arrives, TCP increases the congestion window to 2 and then sends 2 segments, When the 2 acknowledgements of the segments sent out arrives, they each increase the congestion window by one, thereby increasing the congestion window to 4 . therefore it takes   log2N round trips before TCP can send N segments

Answer:

Option b (Port 22) seems to the appropriate choice.

Explanation:

Below seem to be some measure you should take to correct this mistake.

The SSH server also operates on port 22, by default.  

Other choices don't apply to the specified scenario. So that the argument presented above will be appropriate.

The port number would you open on your webserver is Port 22

For better understanding let's explain SSH means.

From the above we can therefore say that the answer The port number would you open on your webserver is Port 22, is correct

Learn more about SSH and port from:

https://brainly.com/question/13054022

Answer:

a) An ARRAY ( 3 )

B) relatively INSENSITIVE ( 7 )

c) a HYBRID ( 4 )

Explanation:

They both used an array based implementation in the standard form and are both relatively insensitive to the order of the data. The main similarity is that each one creates an environment in which a hybrid strategy works more efficiently.

Quadratic selection sort is the sorting system that sorts out the smallest value in an entire list and arrange the list in such a way that the smallest value comes first.

The instructions operate as follows:

a)

JMP AX :

b)

JMP LIST:

c)

JMP NEAR PTR[DI+3]: 

Learn more:

brainly.com/question/20320915

Answer:

A glass of hot water will freeze, of course, but at a faster rate than a glass of normal temperature water

Answer:

When Scheme is a pure functional language, it cannot include the function DISPLAY. Since the pure functional languages will return the same value whenever the same expression is evaluated, and the DISPLAY would be the output and thus it cannot be part of the purely functional language. Thus in the pure functional language we can evaluate the expressions with the same arguments and it returns the same value since there is no state to change.

Explanation:

Answer:a

What are the activities a Database Designer will do if using the SDLC to design a Database [10]

Q1b

List the five main components of Access Database and state their uses [1

Explanation:

Answer:

The answer is "0 to 4294967295"

Explanation:

Given bit:

32 bits

Calculating smallest and largest bits:

In any processor, it works only on the 0 and 1 bit and it also stores 0 and 1 values. so, the smallest address in bit value is= 0 and the largest bit value can be defined as follows:

largest address value in bits:

[tex]\Longrightarrow 2^{32}-1\\\\\Longrightarrow 4294967296 -1\\\\\Longrightarrow 4294967295\\\\[/tex]

smallest address= 0

largest address = 4294967295

Answer: True

Explanation: True

Answer:

Following are the code to this question:

var id= document.createElement('script');//defining id varaible that uses the createElement method  

ld.src = "https://code.jquery.com/jquery-3.5.1.min.js";//defining id varaible with src that holds the source link

alert("The loaded Script is: ");//defining alert box that print Script loaded.

Explanation:

In the above-given code, an id variable is defined that uses the createElement method, which helps to create the specified name with the Element Node, and in the next line, it uses the src with the id variable, that store the link, it also helps to loads the 3rd party script, which is displayed in an alert msg box.

Answer:

false i think.

Explanation:

Gradient Descent is more likely to reach a local minima. because starting at different points and just in general having a different starting point, will lead us to a different local minimum( aka the lowest point closest to the starting point). if alpha(the learning rate) is too large, gradient descent may fail to converge and may even diverge.

Answer:

Here is the Python program:

def input_type(input_string): #function that takes a string as input and returns the type of that string

   try:   #used to test a block of code for input string type              

       if type(eval(input_string)) == float: #if the type of the string is float

           return "float" #function returns float

       elif type(eval(input_string)) == int: #else if the type of the string is int

           return "integer"#function return integers

       elif type(eval(input_string)) == bool:#else if the type of the string is bool

           return "boolean" #function return boolean

   except: #if the type of the string is neither int, bool nor float

       return "string" #function return boolean

#following statements are used to  test the working of the above function

print(input_type(" ")) #prints string

print (input_type ("False")) #prints boolean

print (input_type ("7.432621")) #prints  float

print (input_type("2788"))#prints int

Explanation:

The function input_type() takes a user input and returns the type of that string input the user entered. The user input can be one of the following types:

int which is the data type of an integer for example 1, 2, 3 etc

float which is the data type of a floating point number for example 1.2, 3.4 etc

bool which is the data type of a boolean for example True or False

string that can be characters for example an empty space character " "

try block here is used to generate an exception and it is used to test a block in the code that checks for the type of the user input. The except block here is used to handle that error. This is executed when the try part raises an error.

Inside the try block, the if and elif statements check the type of the string.

if type(eval(input_string)) == float: statement checks if the type of the string is float. Here the eval() method is used to convert the input string because the user input is basically a string value. So this method evaluates the input_string.

type() method is used to return the type of that evaluated user input string. If condition checks if type(eval(input_string)) which is the type of the evaluated input string is equal to float. If this condition evaluates to true then the function returns "float" otherwise it moves to the next elif statement. Similarly the type of the user input is checked for int in elif part and if both these if and elif conditions evaluate to false then the next elif statement is executed that checks the user input to be bool. If all these three if elif statements evaluate to false then the except part executes which returns "string". The program along with its output is attached in a screenshot.    

Answer:

A. True

Explanation:

The correct option is A. True.

Actually, in using cloud infrastructures, the client necessarily cedes control to the CP on a number of issues which may affect security. It may have to restrict port scans and even penetration testing. Moreover, the service level agreements may not actually guarantee commitment to offer such services on the part of the CP. This may result in leaving a gap in security defenses.

Also, when the control of the CP changes, it's evident that the terms and conditions of their services may also change which may affect security.

Answer:

i would check all connections from the hard drive and power supply and check your motherboard for any missing sauters and or loose no contact connections

Explanation:

Answer:

The answer is "secret key k".

Explanation:

When the sender sends the message, he/she may combine, it with the hidden key, and determines the hashing, when transmitting the information the digest is being sent, and also the message is transmitted through the connection.

when the receiver receives the message, still having the very same secret key, adding the Hidden Key then recalculating the hashing again. In this, two factors if another subsequent digest matches something, this message is sent.  

Answer:

?

Explanation:

these are instructions to a question?

Answer:

a)  

int apples [5] = {2, 4, 6, 8, 10};

b)

int *aPtr   //this is the pointer to int

Another way to attach a pointer to a an int variable that already exists:

int * aPtr;

int var;

aPtr = &var;

c)

for (int i = 0; i < size; i++){

       cout << values[i] << endl;    }

d)  

   aPtr = values;

   aPtr = &values[0];    

both  the statements are equivalent

e)

If its referring to the part d) then the address is:

cout<<aPtr;

f)

     for (int i = 0; i < size; ++i) {

            cout<<*(vPtr + i)<<endl;    }

g)

   cout << (aPtr + 3) << endl;  // address referenced by aPtr + 3

   cout << *(aPtr + 3) << endl; // value stored at that location

This value stored at location is 8

h)

    aPtr = &apples[4];

    aPtr -= 4;

    cout<<aPtr<<endl;

    cout<<*aPtr<<endl;  

Explanation:        

a)

int apples [5] = {2, 4, 6, 8, 10};

In this statement the array names is apples, the size of the array is specified in square brackets. so the size is 5. The type of array apples is int this means it can store integer elements. The values or elements of the array apples are even integers from 2 to 10. So the elements of array are:

apples[0] = 2

apples[1] = 4

apples[2] = 6

apples[3] = 8

apples[4] = 10

b)

In this statement int *aPtr  

The int* here is used to make the pointer aPtr points to integer object. Data type the pointer is pointing to is int. The asterisk symbol used with in makes this variable aPtr a pointer.

If there already exists an int type variable i.e. var and we want the pointer to point to that variable then declare an int type pointer aPtr and aPtr = &var; assigns the address of variable var to aPtr.

int * aPtr;

int var;

aPtr = &var;

c)

The complete program is:

int size= 5;

int values[size] = {2,4,6,8,10};

for (int i = 0; i < size; i++){

       cout << values[i] << endl; }

The size of array is 5. The name of array is values. The elements of array are 2,4,6,8,10.

To print each element of the values array using array subscript notation, the variable i is initialized to 0, because array index starts at 0. The cout statement inside body of loop prints the element at 0-th index i.e. the first element of values array at first iteration. Then i is incremented by 1 each time the loop iterates, and this loop continues to execute until the value of i get greater of equal to the size i.e. 5 of values array.

The output is:

2

4

6

8

10

d)

aPtr = values;

This statement assigns the first element in values array to pointer aPtr. Here values is the address of the first element of the array.

aPtr = &values[0];    

In this statement &values[0] is the starting address of the array values to which is assigned to aPtr. Note that the values[0] is the first element of the array values.

e)

Since &values[0] is the starting address of the array values to which is assigned to aPtr. So this address is the physical address of the starting of the array. If referring to the part d) then use this statement to print physical address is aPtr pointing to

cout<<aPtr;

This is basically the starting address of the array values to which is assigned to aPtr.

The output:

0x7fff697e1810                

f)

i variable represents offset and corresponds directly to the array index.

name of the pointer i.e. vPtr references the array

So the statement (vPtr + i) means pointer vPtr that references to array values plus the offset i array index that is to be referenced. This statement gives the address of i-th element of values array. In order to get the value of the i-th element of values array, dereference operator * is used.  It returns an ith value equivalent to the address the vPtr + i is pointing to. So the output is:

2

3

6

8

10

g)

values[0] is stored at 1002500

aPtr + 3 refers to values[3],

An integer is 4 bytes long,

So the address that is referenced by aPtr + 3 is

1002500 + 3 * 4 = 1002512

values[3] is basically the element of values array at 3rd index which is the 4th element of the array so the value stored at that referred location  is 8.

h)

Given that aPtr points to apples[4], so the address stored in aPtr is

1002500 + 4 * 4 = 1002516

aPtr -= 4  is equivalent to aPtr = aPtr - 4

The above statement decrements aPtr by 4 elements of apples array, so the new value is:

1002516 - 4 * 4 = 1002500

This is the address of first element of apples array i.e 2.

Now

cout<<aPtr<<endl; statement prints the address  referenced by aPtr -= 4 which is 1002500  

cout<<*aPtr<<endl;  statement prints the value is stored at that location which is 2.

Answer:

AND

The  AND operator will cause a record to  be selected only if two or more

conditions are satisfied.

Answer:

pie chart

Explanation:

pie chart shows percentage per item

Answer:

pie chart

Explanation:

Answer:

It allows for departmentalization and managerial control. Simple and easy to understand and use. Easy to manage due to the rigidity of the model – each phase has specific deliverables and a review process. Phases are processed and completed one at a time.

Explanation:

Answer:

Complete the program with the following code segment

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

{

cout<<offsetAmount[i]+origList[i]<<" ";

}

return 0;

}

Explanation:

The following line is an iteration of variable i from 1 to 3; It iterates through elements of origList and offsetAmount

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

This adds and prints the corresponding elements of origList and offsetAmount

cout<<offsetAmount[i]+origList[i]<<" ";

}  The iteration ends here

Answer:

Output is the act of the computer display information based on your input

Explanation:

Funny thing is that output is the exact opposite of what you just described

Answer:

The program written in Java is as follows

import java.util.Scanner;

public class mutilate {

public static void main(String [] args) {

Scanner input = new Scanner(System.in);

int n;

System.out.print("Array Length: ");

n = input.nextInt();

while(n<1 || n > 103)  {

    System.out.print("Array Length: ");

n = input.nextInt();

}

int a []= new int [n];

int b []= new int [n];

System.out.print("Enter Elements of the Array: ");

for(int i =0;i<n;i++)  {

 a[i] = input.nextInt();

}

System.out.print("Output: ");

for(int i =0;i<n;i++)  {

 if(i == 0)   {

     b[i] = 0+a[i]+a[i+1];

 }

 else if(i == n-1)   {

     b[i] = a[i - 1]+a[i]+0;

 }

 else   {

 b[i] = a[i - 1]+a[i]+a[i+1];    

 }

 System.out.print(b[i]+" ");

}

}

}

Explanation:

This line allows the program accepts user input

Scanner input = new Scanner(System.in);

This line declares integer n

int n;

The next two line prompts user for length of the array and also accepts input

System.out.print("Array Length: ");

n = input.nextInt();

The following while iteration ensures that the user input is between 1 and 103

while(n<1 || n > 103)  {

    System.out.print("Array Length: ");

n = input.nextInt();

}

The next two lines declares array a and b

int a []= new int [n];

int b []= new int [n];

The next for iteration allows user enter values for array a

System.out.print("Enter Elements of the Array: ");

for(int i =0;i<n;i++)  {

 a[i] = input.nextInt();

}

The next for iteration calculates and prints the values for array b based on the instruction in the question

System.out.print("Output: ");

for(int i =0;i<n;i++)  {

 if(i == 0)   {

     b[i] = 0+a[i]+a[i+1];

 }

 else if(i == n-1)   {

     b[i] = a[i - 1]+a[i]+0;

 }

 else   {

 b[i] = a[i - 1]+a[i]+a[i+1];    

 }

 System.out.print(b[i]+" ");

}