Write a program in your favorite language (e.g., C, Java, C , etc.) to convert max. 32 bit numbers from binary to decimal. The user should type in an unsigned binary number. The program should print the decimal equivalent. You should write a program from scratch that performs the conversion, please do not use built in functions. Please provide a screenshot of your program running on these values: 111011012 110000110011110101110101011101012 Please also include your source code together with your answers to assignment questions.

Answer:there are different ways of quick navigation between files and functions. ... You should use the menu 'Remove file from project' instead of deleting files. ... A Makefile generation tool for Code::Blocks IDE by Mirai Computing

Answer:

Following are the solution to this question:

Explanation:

For point a:

                      [tex]alicerc \ \ \ \ \ \ \ \ \ \ \ bobrc \ \ \ \ \ \ \ \ \ \ \ \ \ \ cyndyrc\\[/tex]

[tex]Alice \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ox \ \ \ \ \ \ \ \ \ \ \ \ \ \ r\\\\ Bob \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ r \ \ \ \ \ \ \ \ \ \ \ \ \ \ ox \\\\Cyndy \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ r\ \ \ \ \ \ \ \ \ \ \ \ \ \ rw \ \ \ \ \ \ \ \ \ \ \ \ \ \ orwx[/tex]

For point b:

               [tex]alicerc \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ bobrc \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ cyndyrc[/tex]

[tex]Alice \ \ \ \ \ \ \ \ \ \ \ \ \ \ ox \ \ \ \ \ \ \ \ \ \ \ \ \ \ r \ \ \ \ \ \ \ \ \ \ \ \ \ \ r\\\\Bob \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ox \\\\Cyndy \ \ \ \ \ \ \ \ \ \ \ \ \ \ r \ \ \ \ \ \ \ \ \ \ \ \ \ \ rw \ \ \ \ \ \ \ \ \ \ \ \ \ \ orwx[/tex]

Answer:

The answer is "1.25"

Explanation:

[tex]{CPU \ time}= {instructions \times CPI \times cycle\ time}[/tex]

[tex]\therefore\\\CPI= \frac{CPU \ time}{instructions \times cycle \ time} \\\\cycle \ time = 1 \\\\ns = 10^{-9} s \\[/tex]

Also for this context, it executes the time =  CPU time. So, the compiler A, we has

[tex]CPI_{A}= \frac{CPU \ time_{A}} {instructions_{A} \times cycle \ time}= \frac{1.1 s}{10^{9} \times 10^{-9} s}= 1.1[/tex]

For compiler B, we have

[tex]CPI_{B}= \frac{CPU \ times_{B}} {instructions_{B} \times cycle \ time}[/tex]

         [tex]= \frac{1.5\ s}{ 1.2 \times 10^{9} \times 10^{-9} \ s}\\\\= \frac{1.5}{ 1.2 }\\\\= 1.25[/tex]

The average CPI for each program given that the processor has a clock cycle time of 1 ns is : 1.1, 1.25.

CPI (Complier A)=CPU time/Instruction×Cycle time

Where:

Cycle time=1ns=10^-9s

Hence:

CPI (Complier A)=1.1s/10^9×10^-9s

CPI (Complier A)=1.1

CPI (Complier B)=CPU time/Instruction×Cycle time

CPI (Complier B)=1.5s/1.2×10^9×10^-9s

CPI (Complier B)=1.25

Inconclusion the average CPI for each program given that the processor has a clock cycle time of 1 ns is : 1.1, 1.25.

Learn more about average CPI here:https://brainly.com/question/24723238

Answer:

doon sa pasay nag kakapi

Answer:

Answer:

At the planning stage maybe?

Explanation:

I'm not positive but in plato it discusses this in the Website Development Proccess lesson.

Answer:

Film and video speak the same audiovisual ’’Technology’’

Answer:

secondary storage

Answer:

What they're saying is that the first two bytes (first two eight bit segments) tell you the width and height of the pattern.

In the example given, you'll notice that the first 16 digits are 00000100 00000100.  If you convert those to decimal, you'll see that those are both equal to four.

If instead the second block of eight bits was 00000111, the image height would then be seven.

Answer:

binary digits in computer system it belongs

Answer:

B

Explanation:

Answer:

I believe the answer is B

Answer:

1.48 s

Explanation:

Number of instructions = 500 million = 500 * 10⁶

clock rate = 1 / 2.2 GHz = 1 / (2.2 * 10⁹ Hz) = 0.4545 * 10⁻⁹ s

We need to compute the clocks per instruction (CPI)

The CPI = summation of (value * frequency)

CPI = (50% * 3 clock cycles) + (50% * 10 clock cycles)

CPI = (0.5 * 3) + (0.5 * 10) = 1.5 + 5 = 6.5

Execution time = number of instructions * CPI * clock rate

Execution time = 500 * 10⁶ * 6.5 * 0.4545 * 10⁻⁹ =1.48 s

Answer:

The function is as follows:

void sendEmail(string name, string prize, string win_lose){

cout << "Dear "<<name<<", " << endl;

cout << "You are the "<<win_lose<<" of our raffle for charity." << endl;

cout << "The prize was: "<<prize<< endl;

cout << "Thank you for giving to charity!" << endl;

cout << "Sincerely," << endl;

cout << "The Boolean Foundation" << endl;

}

Explanation:

This defines the function along the three parameters

void sendEmail(string name, string prize, string win_lose){

This prints the salutation with the person's name

cout << "Dear "<<name<<", " << endl;

This prints if the person won or lost

cout << "You are the "<<win_lose<<" of our raffle for charity." << endl;

This prints the prize, if any

cout << "The prize was: "<<prize<< endl;

The following is the closing remark

cout << "Thank you for giving to charity!" << endl;

cout << "Sincerely," << endl;

cout << "The Boolean Foundation" << endl;

}

Answer:

In Python:

(a) Concatenate strings:

joined = str1+" "+str2

(b) Two Letters in Word :

two_letters = word[4]+word[-3]

(c) Set Numbers in Card

if num_cards < 7:

    num_cards = 7

Explanation:

The code segments were written in Python

All variables were assumed to have been initialized

Solving (a): Concatenate strings:

To do this, we make use of + operator.

So, the concatenation of str1 and str2 with space in between is

str1+" "+str2

When assigned to variable joined, it becomes

joined = str1+" "+str2

Solving (b): Two Letters in Word :

The character at the 5th position is represented with index 4 i.e. word[4]

To access a character from the end, we make use of - sign. So, the third character from the end is word[-3]

Concatenate them using + operator.

So, we have:

two_letters = word[4]+word[-3]

Solving (c): Set Numbers in Card

Here, we make use of the if condtion.

First, check if num_cards is less than 7(i.e. num_cards < 7)

If true, assign num_cards to 7

So, we have:

if num_cards < 7:

    num_cards = 7

Answer:

a

Explanation:

Answer:

Pixel dimensions

Explanation:

measure of height and width of a image in pixels

Answer:

{1, 10, 100}

{a : a ∈ Z and a > 10}

{a : a ∈ N and a < 10}

∅

{ε}

{abc}

Explanation:

1.) A set containing the numbers 1, 10 and 100

2.) Z represents integers, hence numbers n the set are integers values greater Than 100

3.) N represents natural numbers, this the set contains natural numbers less Than 10

4.)∅ represents a null or empty set

5.)represents an empty string

6) contains the sting values a, b and c

Answer:

B. {1, 2, 2, 3, 3, 4, 5}

Explanation:

Given

The above code segment

Required

Determine which list does not work

The list that didn't work is [tex]B.\ \{1, 2, 2, 3, 3, 4, 5\}[/tex]

Considering options (A) to (E), we notice that only list B has consecutive duplicate numbers i.e. 2,2 and 3,3

All other list do not have consecutive duplicate numbers

Option B can be represented as:

[tex]nums[0] = 1[/tex]

[tex]nums[1] = 2[/tex]

[tex]nums[2] = 2[/tex]

[tex]nums[3] = 3[/tex]

[tex]nums[4] = 3[/tex]

[tex]nums[5] = 4[/tex]

[tex]nums[6] = 5[/tex]

if (nums.get(j).equals(nums.get(j + 1)))

The above if condition checks for duplicate numbers.

In (B), when the elements at index 1 and 2 (i.e. 2 and 2) are compared, one of the 2's is removed and the Arraylist becomes:

[tex]nums[0] = 1[/tex]

[tex]nums[1] = 2[/tex]

[tex]nums[2] = 3[/tex]

[tex]nums[3] = 3[/tex]

[tex]nums[4] = 4[/tex]

[tex]nums[5] = 5[/tex]

The next comparison is: index 3 and 4. Meaning that comparison of index 2 and 3 has been skipped.

This is so because of the way the if statement is constructed.

There are different  kinds of ArrayList of integers. The option in the lists can be passed to remDups to show that the method does not work as intended is {1, 2, 2, 3, 3, 4, 5}.

In Python, there are different ways to remove duplicates from list. An example using naive or novice method is;

The main list is : [1, 3, 5, 6, 3, 5, 6, 1]

The list after deleting duplicates : [1, 3, 5, 6]

Learn more about coding from

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Answer:

send and receive messages via the internet

Answer:

a and b

Explanation:

because iam smart

Answer:

logical is the answer

hope it helps

Answer:

logical

Explanation:  keep it real witcha

Explanation:

Hyperlink is the primary method used to navigate between webpages.

Hyperlink can redirect us to another webpages, such as websites that has graphics, files, sounds on the same webpage.

Answer:

weight = int(input("Enter weight in milligrams: "))

kilograms = int(weight / 1000000)

grams = int((weight - (kilograms * 1000000)) / 1000)

milligrams = weight - ((kilograms * 1000000) + (grams * 1000))

print("{} milligrams are equivalent to {} kilogram(s), {} gram(s), and {} milligram(s)".format(weight, kilograms, grams, milligrams))

Explanation:

*The code is in Python.

Ask the user to enter the weight and set it to the variable weight

Calculate the kilograms, divide the weight by 1000000 and cast the result to the int (If the weight is 1050042, kilograms would be 1050042/1000000 = 1)

Calculate the grams, subtract the kilograms from the weight, divide it by 1000 and cast the result to the int (If the weight is 1050042, grams would be int((1050042 - (1 * 1000000)) / 1000) = 50)

Calculate the milligrams, subtract the kilograms and grams from the weight (If the weight is 1050042, milligrams would be 1050042 - ((1 * 1000000) + (50 * 1000)) = 42)

Print the weight, kilograms, grams, and milligrams in the required format

In this exercise we have to use the knowledge of the python language to write the code, so we have to:

The code is in the attached photo.

Some important information informed in the statement that we have to use in the code is:

So to make it easier the code can be found at:

weight = int(input("Enter weight in milligrams: "))

kilograms = int(weight / 1000000)

grams = int((weight - (kilograms * 1000000)) / 1000)

milligrams = weight - ((kilograms * 1000000) + (grams * 1000))

print("{} milligrams are equivalent to {} kilogram(s), {} gram(s), and {} milligram(s)".format(weight, kilograms, grams, milligrams))

See more about python at brainly.com/question/26104476

Answer:

HTML, XML and SGML are the markup languages (hence the "ML" suffix on each of their acronyms).

PHP, ASP and Javascript are scripting languages.

A kind of markup language is HTML. It "marks up" or encapsulates data using HTML tags that specify the data and explain how it will be used on the webpage. Following that, the web browser examines the HTML, which instructs it as to what sections are headings, paragraphs, links, etc.

OOP scripting is used in Java, whereas OOP programming is used in Java. JavaScript code can only be run on a browser, while Java applications can run on a virtual machine or browser.

While JavaScript code is completely in text, Java code must be compiled. They both need various plug-ins.

Therefore, The "ML" suffix on each acronym stands for markup languages, which are HTML, XML, and SGML respectively. Scripting languages include JavaScript, ASP, and PHP.

Learn more about languages here:

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#SPJ2

Answer:

Answer:

# global variable for logged in users

loggedin = []

# create the user account list as a dictionary called "bank".

def import_and_create_dictionary(filename):

   bank = {}

   i = 1

   with open(filename, "r") as file:

       for line in file:

           (name, amt) = line.rstrip('\n').split(":")

           bank[i] = [name.strip(), amt]

           i += 1

   return bank

#create the list of user login details.

def import_and_create_accounts(filename):

   acc = []

   with open(filename, "r") as file:

       for line in file:

           (login, password) = line.rstrip('\n').split("-")

           acc.append([login.strip(), password.strip()])

   return acc

# function to login users that are yet to login.

def signin(accounts, login, password):

   flag="true"

   for lst in accounts:

       if(lst[0]==login and lst[1] == password):

           for log in loggedin:

               if(log ==login):

                    print("You are already logged in")

                    flag="false"

           break                    

       else:

           flag="false"  

   if(flag=="true"):

       loggedin.append(login) # adding user to loggedin list

       print("logged in succesfully")

       return True

   return False

# calling both the function to create bank dictionary and user_account list        

bank  = import_and_create_dictionary("bank.txt")            

print (bank)

user_account = import_and_create_accounts("user.txt")

print(user_account)

# check for users loggedin

signin(user_account,"Brandon","123abcAB")

signin(user_account,"James","100jamesABD")

signin(user_account,"Sarah","sd896ssfJJH")

Explanation:

The python program is a banking application system that creates a list of user accounts and their login details from the bank.txt and user.txt file. The program also uses the information to log in and check for users already logged into their account with the "loggedin" global variable.

In this exercise we have to use the knowledge of programming in the python language, in this way we find that the code can be written as:

The code can be seen in the attached image.

To make it even simpler to visualize the code, we can rewrite it below as:

def import_and_create_dictionary(filename):

  bank = {}

  i = 1

  with open(filename, "r") as file:

      for line in file:

          (name, amt) = line.rstrip('\n').split(":")

          bank[i] = [name.strip(), amt]

          i += 1

  return bank

#create the list of user login details.

def import_and_create_accounts(filename):

  acc = []

  with open(filename, "r") as file:

      for line in file:

          (login, password) = line.rstrip('\n').split("-")

          acc.append([login.strip(), password.strip()])

  return acc

def signin(accounts, login, password):

  flag="true"

  for lst in accounts:

      if(lst[0]==login and lst[1] == password):

          for log in loggedin:

              if(log ==login):

                   print("You are already logged in")

                   flag="false"

          break                  

      else:

          flag="false"  

  if(flag=="true"):

      loggedin.append(login) # adding user to loggedin list

      print("logged in succesfully")

      return True

  return False

bank  = import_and_create_dictionary("bank.txt")      

print (bank)

user_account = import_and_create_accounts("user.txt")

print(user_account)

signin(user_account,"Brandon","123abcAB")

signin(user_account,"James","100jamesABD")

signin(user_account,"Sarah","sd896ssfJJH")

See more about programming at brainly.com/question/11288081

Answer:

the older versions and really all of them are

Answer:

This is what I found on the internet that I hope will help you! :3

Explanation:

It's a free app that will be preinstalled with Windows 10, and you don't need an Office 365 subscription to use it. The existing My Office app has many of these features, but the new Office app puts the focus on the free online versions of Office if you're not an Office 365 subscriber.

Answer:

Following are the solution to the given question:

Explanation:

Please find the complete and correct question in the attachment file.  

For Point a:

For the second round,

A is selects kA(2)  either 0 or 1, so for each of them, that is [tex]\frac{1}{2}[/tex].

B selects [tex]kB(2)\ from\ (0, 1, 2, 3)[/tex] for each choice with the probability of [tex]\frac{1}{4}[/tex].

If [tex]kA(2) < kB(2)[/tex] wins the second rear race.

[tex]\to P[A \ wins] = P[kA(2) < kB(2)][/tex]

                     [tex]= P[kA(2) = 0] \times P[kB(2) > 0] + P[kA(2) = 1] \times P[kB(2) > 1]\\\\= \frac{1}{2} \times \frac{3}{4} + \frac{1}{2} \times \frac{2}{4} \\\\=\frac{3}{8} +\frac{2}{8} \\\\= \frac{3+2}{8}\\\\= \frac{5}{8}[/tex]

For Point b:

Throughout this example, [tex]kA(3)[/tex] also selects to be either 0 or 1 with such a [tex]\frac{1}{2}[/tex] probability. So, although B chooses [tex]kB(3)[/tex] from [tex](0, 1, 2, 3, 4, 5, 6, 7),[/tex] the probabilities each are [tex]\frac{1}{8}[/tex]:

[tex]\to P[A \ wins] = P[kA(3) < kB(3)][/tex]

                     [tex]= P[kA(3) = 0] \times P[kB(3) > 0] + P[kA(3) = 1] \times P[kB(3) > 1]\\\\= \frac{1}{2} \times \frac{7}{8} + \frac{1}{2} \times \frac{6}{8}\\\\= \frac{7}{16} + \frac{6}{16}\\\\= \frac{7+6}{16} \\\\= \frac{13}{16}\\\\[/tex]

For point c:

Assume that B tries again 16 times (typical value), and it destroys. In addition, throughout the exponential background n is obtained at 10 when choosing k between 0 to 2n−1. The probability of A winning all 13 backoff events is: [tex]P[A \text{wins remaining races}] = 16\pi i =4P[A \ wins \ i |A \ wins \ i -1 ][/tex]

Let the k value kA(i) be A for the backoff race I select. Because A retains the breed

[tex]=(kA(i)] \cdot P[kA(i+ 1)< kB(i+ 1)] \geq P[kA(i) + 1<kB(i)] \cdot P[kA(i+ 1)< kB(i+1)]+P[kA(i) + 1 \geq kB(i)] \cdot P[kA(i+ 1)< kB(i+ 1)] \\\\= (P[kA(i) + 1< kB(i)] +P[kA(i) + 1 \geq kB(i)]) \times P[kA(i+ 1) < kB(i+ 1)]\\\\=P[kA(i+ 1)< kB(i+ 1)]\\\\[/tex]

For point d:

Two stations A and B are supposed. They assume that B will try 16 times afterward. Even so, for A, 16 races were likely to also be won at a rate of 0.82 For Just higher expectations of three A, B, and C stations. For Station A, possibility to win all backoffs