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swat32
4 years ago
10

Why would businesses apply cell protection to spreadsheet entries? Provide one specific example.

Computers and Technology
1 answer:
docker41 [41]4 years ago
6 0

Answer:

To protect a formula

Explanation:

One common example to apply cell protection to spreadsheet entries is to protect a formula used in the sheet to calculate payouts or rankings for example.   A manager might want to share the results of the team to all its team and provides some ranking or other form of calculations in the sheet.  He then needs to protect the formula so it's not altered by the team members or anyone else reviewing the file.

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Explanation:

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3 years ago
The process of __________ encourages members to accept responsibility for the outcomes of a group and for changing the style in
soldier1979 [14.2K]

Answer:

The correct answer to the following question will be Option B (Interpersonal feedback).

Explanation:

  • Interpersonal feedback seems to be a mechanism under which someone talks. Individuals deliver a message regarding their sensations and expectations of actions by another.
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Therefore, Interpersonal feedback is the right answer.

8 0
3 years ago
Purpose of this project is to increase your understanding of data, address, memory contents, and strings. You will be expected t
STALIN [3.7K]

Answer:

See explaination for the details

Explanation:

#Starting point for code/programm

main:

la $a0,newLine #Print a new line

li $v0,4

syscall

# Find the number of occurence of a string in the given sentence

la $a0,prompt1 # Prompt the user to enter the first string.

li $v0,4

syscall

li $v0, 8 # Service 8 = read input string

la $a0, fword

li $a1, 9

syscall

la $a0,prompt2 # Prompt the user to enter the second string.

li $v0,4

syscall

li $v0, 8 # Service 8 = read input string

la $a0, sword

li $a1, 9

syscall

# process first word

li $t4,0 # Intialize the couter to 0

la $t0,sstatement # Store the statement into $t0

nstart1: la $t1,fword # Store the search word into $t1

loop1: # loop1 finds the number of occurences

# of input word in the given statment

lb $t2,($t0) # Load the starting address(character) of

# sstatement into $t2

lb $t3,($t1) # Load the starting address of input word

# into $t3

beq $t3,'\n',inc_counter1

beqz $t3,inc_counter1 # If $t3 is null , exit loop and print output

beqz $t2,print_output1 # If $t2 is null , exit loop and print output

move $a0,$t2 # Convert $t2 to lower, if it is upper case

jal convert2lower

move $t2,$v0 # Store the return($v0) value into $t2

move $a0,$t3 # Convert $t3 to lower, if it is upper case

jal convert2lower

move $t3,$v0 # Store the return($v0) value into $t3

bne $t2,$t3,next_char1 # If both characters are not matched current

# character in the string, go to next character

addiu $t0,$t0,1 # otherwise, increment both indexes

addiu $t1,$t1,1

j loop1 # go to starting of the loop

next_char1:

la $t5,fword

bne $t5,$t1,nstart1

la $t1,fword # Store the input word into $t1

addiu $t0,$t0,1 # Increment the index to goto next character

j loop1 # go to starting of the loop

inc_counter1:

addi $t4,$t4,1 # Increment the frequency counter by 1

la $t1,fword # Store input word into $t1

j loop1 # go to starting of the loop

print_output1:

la $t0,fword

L1:

lb $a0,($t0)

beq $a0,'\n',exL1

jal convert2upper

move $a0,$v0

li $v0,11

syscall

addiu $t0,$t0,1

j L1

exL1:

la $a0,colon

li $v0,4

syscall

la $a0, dash

li $v0, 4

syscall

move $a0,$t4

li $v0,1

syscall # print new line

la $a0,newLine

li $v0,4

syscall

# process second word

li $t4,0 # Intialize the couter to 0

la $t0,sstatement # Store the statement into $t0

nstart2: la $t1,sword # Store the search word into $t1

loop2: # loop1 finds the number of occurences

# of input word in the given statment

lb $t2,($t0) # Load the starting address(character) of

# sstatement into $t2

lb $t3,($t1) # Load the starting address of input word

# into $t3

beq $t3,'\n',inc_counter2

beqz $t3,inc_counter2 # If $t3 is null , exit loop and print output

beqz $t2,print_output2 # If $t2 is null , exit loop and print output

move $a0,$t2 # Convert $t2 to lower, if it is upper case

jal convert2lower

move $t2,$v0 # Store the return($v0) value into $t2

move $a0,$t3 # Convert $t3 to lower, if it is upper case

jal convert2lower

move $t3,$v0 # Store the return($v0) value into $t3

bne $t2,$t3,next_char2 # If both characters are not matched current

# character in the string, go to next character

addiu $t0,$t0,1 # otherwise, increment both indexes

addiu $t1,$t1,1

j loop2 # go to starting of the loop

next_char2:

la $t5,sword

bne $t5,$t1,nstart2

la $t1,sword # Store the input word into $t1

addiu $t0,$t0,1 # Increment the index to goto next character

j loop2 # go to starting of the loop

inc_counter2:

addi $t4,$t4,1 # Increment the frequency counter by 1

la $t1,sword # Store input word into $t1

j loop2 # go to starting of the loop

print_output2:

la $t0,sword

L2:

lb $a0,($t0)

beq $a0,'\n',exL2

jal convert2upper

move $a0,$v0

li $v0,11

syscall

addiu $t0,$t0,1

j L2

exL2:

la $a0,colon

li $v0,4

syscall

la $a0, dash2

li $v0, 4

syscall

move $a0,$t4

li $v0,1

syscall

exit:

# Otherwise, end the program

li $v0, 10 # Service 10 = exit or end program

syscall

############################ subroutine - convert2lower #################################

convert2lower: # Converts a character(stored in $a0) to

# its lower case, if it is upper case

# and store the result(lower case) in $v0

move $v0,$a0

blt $a0,'A',return

bgt $a0,'Z',return

subi $v0,$a0,-32

return: jr $ra # Return the converted(lower case) character

############################## subroutine - convert2upper ##################################

convert2upper: # Converts a character(stored in $a0) to

# its upper case, if it is lower case

# and store the result(upper case) in $v0

move $v0,$a0

blt $a0,'a',return2

bgt $a0,'z',return2

addiu $v0,$a0,-32

return2: jr $ra # Return the converted(lower case) character

4 0
3 years ago
Which of the following is preventing the remote user from being able to access the workstation?
larisa86 [58]

Answer:

Option C is the correct option to the following question.

Explanation:

In the following statement, the absence of the network time simultaneous occurrence is generated by the authentication mismatches then, the remote user is preventing access to the workstations. So that's why the network time is absent that caused the prevention in the remote user by accessing the workstations.

5 0
4 years ago
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