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murzikaleks [220]
3 years ago
11

It is necessary tto save updates often when working in google docs? True or false

Computers and Technology
2 answers:
borishaifa [10]3 years ago
7 0
True cause sometimes it can delete all ur work
kiruha [24]3 years ago
7 0

Answer:

false

Explanation:

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OSI layer for HDLC??​
tresset_1 [31]

Answer:

HDLC is one of the most commonly used internet protocols (IP) in what is Layer 2 of the industry communication reference model called Open Systems Interconnection (OSI).

Explanation:

3 0
3 years ago
A _ shows the end of a page . ​
marta [7]

Answer:

page break shows the end of a page.

4 0
2 years ago
Read 2 more answers
python A pet shop wants to give a discount to its clients if they buy one or more pets. The discount is equal to 20 percent of t
Artyom0805 [142]

Answer:

The following are the program in the Python Programming Language.

#define function

def discount(prices, isPet, nItems):

 # declare and initialize variables to 0

 i = 0

 cost = 0

 pets = 0

 item = 0

 item_Cost = 0

 #set the while loop

 while(i < nItems):

   #check the items of the variable

   if isPet[i]:

     #add and initialize in the cost

     cost += prices[i]

     #increament in the variable by 1

     pets += 1

   #otherwise

   else:

     #add and initialize in the item_Cost

     item_Cost += prices[i]

     #increament in the variable by 1

     item += 1

   #then increment in the variable by 1

   i += 1

 #check the pet is greater than equal to 1  

 #and item is greater than equal to 5

 if(pets >= 1 and item >=5):

   #then, print the message

   print("You receive discount")

   #set discount of the items

   discount = 0.2 * item_Cost

   # calculate bill after deducting discount

   bill = cost + item_Cost - discount

   #print the bill

   print("Final Bill amount is", bill)

 #otherwise

 else:

   #print the message

   print("You do not receive any discount")

   #calculate bill without deducting discount

   bill = cost + item_Cost

   #print the final bill

   print("Final Bill amount is", bill)

#set empty list type variables

price = []

pet = []

#set the infinite loop

while True:

 #get input from the user

 sprice = int(input("Enter the price (-1 to quit): "))

 #check that price is not equal to the -1

 if(sprice != -1):

   #add price in the list

   price.append(sprice)

   #ask for choice from the user

   choice = input("Is it a pet (Y / N)? ")

   #check the choice of the user is y or Y

   if(choice == 'Y' or choice == 'y'):

     # then, add in the pet

     pet.append(True)

   #otherwise

   else:

     #not add in the pet

     pet.append(False)

   print("")

 #break the loop

 else:

   break

#declare variable that store the length

items = len(price)

#call and initialize the variable

discount(price, pet, items)

<u>Output</u>:

Enter the price (-1 to quit): 96

Is it a pet (Y / N)? y

Enter the price (-1 to quit): 69

Is it a pet (Y / N)? n

Enter the price (-1 to quit): 41

Is it a pet (Y / N)? n

Enter the price (-1 to quit): 52

Is it a pet (Y / N)? n

Enter the price (-1 to quit): 96

Is it a pet (Y / N)? n

Enter the price (-1 to quit): 74

Is it a pet (Y / N)? n

Enter the price (-1 to quit): -1

You receive discount

Final Bill amount is 361.6

Explanation:

<u>The following are the description of the program</u>.

  • Firstly, define the function 'discount()' and pass the arguments 'prices', 'isPet' and 'nItems' in its parameter.
  • Then, declare and initialize the variables 'i' to 0, 'item_Cost' to 0
  • , item to 0, 'cost' to 0 and 'pets' to 0.
  • Then, set the while loop to add the price of the items and the cost of the pets.
  • Set the if-else conditional statement to check the pet is greater than equal to 1 and the items are greater than equal to 5 then, set the discount and print the bill after deduction.
  • Otherwise, print the bill without deduction.
  • Finally, set two list type variable then, set the infinite while loop that gets input from the user and check that the variable is not equal to -1 then append the data in the list then again check the user input then append in pet otherwise not append in pet and break the loop. Then, set the variable that stores the length of the price and call the following function by passing arguments in its parameter.
8 0
3 years ago
What are the layers in the internet protocol stack? what are the principle responsibilities of each of these layers?
Andrew [12]
1: Application
2: Presentation
3:Session
4: Transport
5:Network 
6: Data link 
7: Physical

Application is a Interface Layer and responsible Layer
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3 0
3 years ago
Use semaphore(s) to solve the following problem. There are three processes: P1, P2, and P3. Each process Pi has a segment of cod
Lisa [10]

Answer:

See explaination

Explanation:

Here we will use two semaphore variables to satisfy our goal

We will initialize s1=1 and s2=1 globally and they are accessed by all 3 processes and use up and down operations in following way

Code:-

s1,s2=1

P1 P2 P3

P(s1)

P(s2)

C1

V(s2) .

P(s2). .

. C2

V(s1) .

P(s1)

. . C3

V(s2)

Explanation:-

The P(s1) stands for down operation for semaphore s1 and V(s1) stands for Up operation for semaphore s1.

The Down operation on s1=1 will make it s1=0 and our process will execute ,and down on s1=0 will block the process

The Up operation on s1=0 will unblock the process and on s1=1 will be normal execution of process

Now in the above code:

1)If C1 is executed first then it means down on s1,s2 will make it zero and up on s2 will make it 1, so in that case C3 cannot execute because P3 has down operation on s1 before C3 ,so C2 will execute by performing down on s2 and after that Up on s1 will be done by P2 and then C3 can execute

So our first condition gets satisfied

2)If C1 is not executed earlier means:-

a)If C2 is executed by performing down on S2 then s2=0,so definitely C3 will be executed because down(s2) in case of C1 will block the process P1 and after C3 execute Up operation on s2 ,C1 can execute because P1 gets unblocked .

b)If C3 is executed by performing down on s1 then s1=0 ,so definitely C2 will be executed now ,because down on s1 will block the process P1 and after that P2 will perform up on s1 ,so P1 gets unblocked

So C1 will be executed after C2 and C3 ,hence our 2nd condition satisfied.

4 0
3 years ago
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