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ankoles [38]
3 years ago
9

When you turn on a computer, Windows starts and displays a ____ consisting of the time and date.

Computers and Technology
2 answers:
nata0808 [166]3 years ago
7 0

It really depends what device you're on.


For example, my wifes laptop shows a lock screen, then it shows a sign in screen.

My laptop shows ONLY the sign in screen.


My old laptop shows a sign in screen with the date and time. So I dunno man

ludmilkaskok [199]3 years ago
6 0

The sign in screen shows the time and date and also the textbooks for your username and password. I don't think Windows has a "username screen".

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Svetllana [295]

Answer:

She's procrastinating

Explanation:

3 0
3 years ago
A year in the modern Gregorian Calendar consists of 365 days. In reality, the earth takes longer to rotate around the sun. To ac
Vikentia [17]

Answer:

input-year taken 2020

2020 is divisible by 2

output- 2020 is a leap year

3 0
3 years ago
I need major help. Word processing
Pie

Answer:

I do Word processing in computing, but I use a computing textbook for studies.

I also don't have any idea about the flashcards.

I'm sorry.

Explanation:

But if you want me to help, I may.

3 0
2 years ago
Write a function called has_duplicates that takes a string parameter and returns True if the string has any repeated characters.
vichka [17]

Answer:

Explanation:

The python code for the question is attached in the image below.

The objective here is to write a code in python with a function called has_duplicates that takes a string parameter and returns True if the string has any repeated characters. Otherwise, it should return False.

SEE BELOW FOR THE CODE.

alphabet = "abcdefghijklmnopqrstuvwxyz"

test_dups = ["zzz","dog","bookkeeper","subdermatoglyphic","subdermatoglyphics"]

test_miss = ["zzz","subdermatoglyphic","the quick brown fox jumps over the lazy dog"]

def histogram(s):

   d = dict()

   for c in s:

       if c not in d:

           d[c]=1

       else:

           d[c] += 1

   return d

def has_duplicates():

   for string in test_dups:

       dictionary=histogram(string)

       duplicates=False

       for ch in dictionary.keys():

           if dictionary[ch] > 1:

               duplicates=True

               break

       if duplicates==True:

           print(string,"has duplicates")

       else:

           print(string,"has no duplicates")

def missing_letters(string):

   answer=""

   for ch in alphabet:

       if ch not in string:

           answer=answer+ch

   return answer

print("__________________________________________")

print(" Calling has_duplicates() function")

print("__________________________________________")

has_duplicates()

print("\n______________________________________________")

print("Calling missing_letters() function in for loop")

print("______________________________________________")

for i in test_miss:

   answer=missing_letters(i)

   if len(answer)>=1:

       print(i,"is missing letters",answer)

   else:

       print(i,"uses all the letters")

3 0
3 years ago
Half of the integers stored in the array data are positive, and half are negative. Determine the
jonny [76]

Answer:

Check the explanation

Explanation:

Below is the approx assembly code for above `for loop` :-

1). mov ecx, 0

2). loop_start :

3).    cmp ecx, ARRAY_LENGTH

4).    jge loop_end

5).    mv temp_a, array[ecx]

6).    cmp temp_a, 0

7).    branch on nge

8).        mv array[ecx], temp_a*2

9).   add ecx, 1

10).   jmp loop_start

11). loop_end :

Assumptions :-

*ARRAY_LENGTH is register with value 1000000

*temp_a is a register

Frequency of statements :-

1) will be executed one time

3) will be executed 1000000 times

4) will be executed 1000000 times

5) will be executed 1000000 times

6) will be executed 1000000 times

7) `nge` will be executed 1000000 times, branch will be executed 500000 times

8) will be executed 500000 times

9) will be executed 1000000 times

10) will be executed 1000000 times

Cost of statements :-

1) 10 ns

3) 10ns + 10ns + 10ns [for two register accesses and one cmp]

4) 10ns [for jge ]

5) 10ns + 100ns + 10ns [10ns for register access `ecx`, 100ns for memory access `array[ecx]`, 10ns for mv]

6) 10ns + 10ns [10ns for register_access `temp_a`, 10ns for mv]

7) 10ns for nge, 10ns for branch

8) 30ns + 110ns + 10ns

10ns + 10ns + 10ns for temp_a*2 [10ns for moving 2 into a register, 10ns for multiplication],

110ns for array[ecx],

10ns for mv

9) 10ns for add, 10ns for `ecx` register access

10) 10ns for jmp

Total time taken = sum of (frequency x cost) of all the statements

1) 10*1

3) 30 * 1000000

4) 10 * 1000000

5) 120 * 1000000

6) 20 * 1000000

7) (10 * 500000) + (10 * 1000000)

8) 150 * 500000

9) 20 * 1000000

10) 10 * 1000000

Sum up all the above costs, you will get the answer.

It will equate to 0.175 seconds

7 0
3 years ago
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