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balandron [24]
3 years ago
14

How do professionals address their problems?

Computers and Technology
2 answers:
a_sh-v [17]3 years ago
7 0
B) Grace and maturity is the most professional way you could address problems
8090 [49]3 years ago
6 0

Answer:

b, with grace and maturity

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An organization is implementing a preselected baseline of security controls, but finds not all of the controls apply. What shoul
WINSTONCH [101]

The answer is Tailoring the baseline to their needs.

After an organization selects applicable security control baseline but finds not all of the controls apply, it initiates a tailoring process to modify the controls appropriately and more closely with specific conditions related to organizational missions, information systems or environments of operation. It is an integral part of the security control selection and specification and involves risk management processes like assessing, responding to, and framing.

8 0
4 years ago
Design a hierarchy of classes, where the Media superclass has the artistName and the mediaName as the common attributes. Create
zaharov [31]

Answer:

See attached file.

Explanation:

See attached file for detailed code.

Download txt
7 0
3 years ago
What modifier should you use on the members of a class so that they are not accessible to another class in a different package,
vova2212 [387]

The private modifier specifies that only other members of the same class can access the member. The protected modifier restricts access to the member to within its own package.

<h3>What are public and private modifiers?</h3>

The public access modifier permits code from both inside and outside the class to access the class's methods and properties, whereas the private modifier prevents outside code from accessing the class's methods and properties.

Thus, private modifier is the member.

For more details about private modifier, click here

brainly.com/question/15108837

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5 0
2 years ago
Using MARS/MIPS
slega [8]

Answer:

Explanation:

MIPS program which increments from 0 to 15 and display results in Decimal on the console

In this program the user defined procedures print_int and print_eot were used to print the integer values and new line characters(\n) respectively.the mechanisam of the loop is explaine in the comment section of the program.

     addi $s0, $0, 0

     addi $s1, $0, 15

print_int:

              li $v0, 1                # system call to print integer

              syscall                

              jr $ra                     # return

print_eol:                      # prints "\n"

            li $v0, 4            

              la $a0, linebrk      

              syscall              

              jr $ra                  # return

main: . . .          

              li $a0, 0                # print 0

              jal print_int         # print value in $a0

loop:   move $a0, $s0           # print loop count

       jal print_int        

       jal print_eol           # print "\n" character

       addi $s0, $s0, 1        # increment loop count by 1

       ble $s1, $s0, loop      # exit if $s1<$s0

beq $s0, $0, end

end:

MIPS progam to increment from 0 to 15 and display results in Hexadecimal on the console

this program is slightly differed from the previous program in this program the system call issued in print_int is implemented with a system call that prints numbers in hex.

addi $s0, $0, 15

     addi $s1, $0, 0

print_int:

   li      $v0,34                  # syscall number for "print hex"

   syscall                         # issue the syscall

              jr $ra                     # return

print_eol:                      # prints "\n"

            li $v0, 4            

              la $a0, linebrk      

              syscall              

              jr $ra                  # return

main: . . .          

              li $a0, 0                # print 0

              jal print_int         # print value in $a0

loop:   move $a0, $s0           # print loop count

       jal print_int        

       jal print_eol           # print "\n" character

       addi $s0, $s0, 1        # increment loop count by 1

       ble $s1, $s0, loop      # exit if $s0>$s1

beq $s0, $0, end

end:

5 0
3 years ago
Which of the following are not valid assignment statements? A) total = 9;
never [62]

Answer:

The correct option for the given question is option(b) i.e 72 = amount;

Explanation:

In the option(a) the total variable store the value 9 which is a valid assignment  

In the option(b) 72 is not a variable because the variable will never start with a numeric number. Therefore it is not a valid assignment statement.

In the option(c) the yourAge variable store the value of myAge variable which is a valid assignment.

So correct answer is an option(b).

7 0
3 years ago
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