What bug was supposed to start affecting computers on january 1 2000

Given that a function receives three parameters a, b, c, of type double , write some code, to be included as part of the functio

hoa [83]

Answer:

public static void quad(double a, double b, double c) {

double D = (Math.pow(b, 2)) - (4 * a * c);

if (D<0){

System.out.println("no real solutions");

}

Explanation:

The code snippet above implements the function in Java programming language

As required by the question, the line double D = (Math.pow(b, 2)) - (4 * a * c); evaluates b squared" - 4ac and assignes the value to the variable D

An if statement is used to test if D is less than 0, if this is true the message no real solutions is printed

5 0

3 years ago

Which design principle is the subject of a photo or image?

VashaNatasha [74]

The focal point is where the eye is drawn to in a photo or image.

6 0

3 years ago

which of the following present the most significant obstacle to developing IT security if we consider commerce and marketing?

Dmitry_Shevchenko [17]

Answer:

There is no direct return on investment in building security systems.

Security systems are detrimental to usability and can make IT systems less functional, and therefore less attractive to the customer.

There is pressure to reduce the time it takes to get a new IT product or system onto the market, so security systems are sacrificed in order to reduce the time-to-market.

Explanation:

Cyber security has always been challenging for the organizations. There have been groups which are always looking for loop holes in cyber security and hacks the details and asks the ransom to restore. IT systems have now been more complex in this era. Users are increasing every new day and network accounts security is more demanding. Computer connection are more complex and require special attention to control them. The obstacles in IT systems are of concern as there is need for dynamic IT solution to counter the challenging hackers. New programs and customized demand of IT systems need customized IT security systems.

4 0

2 years ago

Program:

Mkey [24]

Answer:

See explaination

Explanation:

#method to print menu & handle user choice

def print_menu(input_str):

#printing menu

print('MENU')

print('c - Number of non-whitespace characters')

print('w - Number of words')

print('f - Fix capitalization')

print('r - Replace punctuation')

print('s - Shorten spaces')

print('q - Quit\n')

#getting choice

choice=input('Choose an option: ').lower()

#identifying choice

if choice=='c':

#displaying number of non white space chars in input_str

print('Number of non-whitespace characters:',get_num_of_non_WS_characters(input_str))

elif choice=='w':

#displaying number of words in input_str

print('Number of words:',get_num_of_words(input_str))

elif choice=='f':

#fixing capitalization and getting updated string and count of values capitalized

input_str,count=fix_capilization(input_str)

#displaying results

print('Number of letters capitalized:',count)

print('Edited text:',input_str)

elif choice=='r':

#replacing punctuation, displaying updated text

input_str=replace_punctuation(input_str)

print('Edited text:', input_str)

elif choice=='s':

#shortening spaces and displaying updated text

input_str = shorten_space(input_str)

print('Edited text:', input_str)

#returning choice and input_str

return choice,input_str

#returns the number of non white space chars in input_str

def get_num_of_non_WS_characters(input_str):

count=0

#looping through each character in input_str

for i in input_str:

if not i.isspace():

#i is not a space

count+=1

return count

#returns the number of words in input_str

def get_num_of_words(input_str):

#splitting words into list of tokens by space

words=input_str.split(' ')

count=0

#counting all non empty strings in words list

for i in words:

if len(i)>0:

count+=1

return count

#method to fix capitalization and return updated string and count of letters updated

def fix_capilization(input_str):

count=0

beginning=True #starting letter should be capitalized

result=''

for i in input_str:

if beginning and i.isalpha():

#start of a sentence and i is alphabetic

if i.islower():

#converting i to upper case, incrementing count

i=i.upper()

count+=1

result+=i

#not start of a sentence

beginning=False

elif i in '?.!':

#i is either ? or . or !, next letter should be capitalized

beginning=True

result+=i

else:

#any other character

result+=i

return result,count

#method to replace exclamation and semicolons with period and comma respectively

def replace_punctuation(input_str,exclamationCount =0,semicolonCount=0):

result=''

for i in input_str:

if i=='!':

i='.'

exclamationCount+=1

elif i==';':

i=','

semicolonCount+=1

result+=i

print('Punctuation replaced')

#displaying replaced values counts

print('exclamationCount:',exclamationCount)

print('semicolonCount:',semicolonCount)

return result

#removes all double or more spaces in input_str

def shorten_space(input_str):

input_str=input_str.strip()

result=''

prev=None

for i in input_str:

if prev==None:

result+=i

elif i==' ':

if prev != ' ':

result+=i

else:

result+=i

prev=i

return result

if __name__ == '__main__':

#getting input, printing it

input_str=input('Enter a sample text:\n')

print('\nYou entered:',input_str)

choice=' '

#looping until choice becomes q

while choice!='q':

choice,input_str=print_menu(input_str)

7 0

3 years ago

What is a secondary storage medium that uses magnetic techniques to store and retrieve data on disks or tapes coated with magnet

Elza [17]

Answer:

The correct answer to the following question will be "Magnetic medium".

Explanation:

Some storage device that portrays details or information using magnetic patterns is called a magnetic device.

Magnetic tape, and hard drives are the commonly used magnetic storage devices.
This will be the method by which magnetic methods are used to hold and retrieve information on tapes.

So, it's the right answer.

3 0

3 years ago