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zloy xaker [14]
3 years ago
5

Paragraph: Read the following two e-mail messages. In three to five sentences, explain why E-mail B is the more appropriate work

place e-mail. Remember to consider netiquette guidelines in your response. E-mail A: DUDE! You totally shorted me on my paycheck! I seriously am upset. It is majorly less than it shold be. I totally worked overtime last pay period!!!! YOU NEED TO FIX THIS RIGHT AWAY. I am going shopping this weekend and i need my cash. E-mail B: When I got my paycheck, I noticed that it was less than I expected. It looks like I did not receive overtime pay for the extra hours I worked last pay period. Can you please review this and let me know how I can get this corrected?
Computers and Technology
1 answer:
exis [7]3 years ago
4 0
E-mail B is the more appropriate workplace e-mail because it’s straightforward, polite, and professional. E-mail A was more accusatory and aggressive, and many people don’t like when they’re being yelled at or accused of something and, as a result, this person may start to dislike that coworker. However, the second e-mail was more polite and made the receiver think of them as a respectful person so that they’re happy to help them out.
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When you add encryption to a powerpoint presentation what does it do
abruzzese [7]

Answer: your document will be inaccessible

5 0
2 years ago
1. Write programming code in C++ for school-based grading system. The program should accept the Subject and the number of studen
satela [25.4K]

In this program, I am using the school-based grading system and the program should accept the subject and the number of students.

Program approach:-

  • Using the necessary header file.
  • Using the standard I/O namespace function.
  • Define the main function.
  • Declare the variable.
  • Display enter obtain marks in 5 subjects.
  • Return the value.

Program:-

//header file

#include<iostream>

//using namespace

using namespace std;

//main method

int main()

{

//declare variable

  int j;

  float mark, sum=0, a;

//display enter obtain marks in 5 subjects

  cout<<"Enter Marks obtained in 5 Subjects: ";

  for(j=0; j<5; j++)

  {

      cin>>mark;

      sum = sum+mark;

  }

  a = sum/5;

//display grade

  cout<<"\nGrade = ";

  if(a>=91 && a<=100)

//display a1

      cout<<"a1";

  else if(a>=81 && a<91)

//display a2

      cout<<"a2";

  else if(a>=71 && a<81)

      cout<<"b1";

  else if(a>=61 && a<71)

      cout<<"b2";

  else if(a>=51 && a<61)

//display c1

      cout<<"c1";

  else if(a>=41 && a<51)

//display c2

      cout<<"c2";

  else if(a>=33 && a<41)

//display d

      cout<<"d";

  else if(a>=21 && a<33)

//display e1

      cout<<"e1";

  else if(a>=0 && a<21)

//display e2

      cout<<"e2";

  else

//display invalid

      cout<<"Invalid!";

  cout<<endl;

//return the value

  return 0;

}

Learn more grading system

brainly.com/question/24298916

7 0
2 years ago
Assume that a large number of consecutive IP addresses are available starting at 198.16.0.0 and suppose that two organizations,
Fiesta28 [93]

Answer & Explanation:

An IP version 4 address is of the form w.x.y.z/s

where s = subnet mask

w = first 8 bit field, x = 2nd 8 bit field, y = 3rd 8 bit field, and z = 4th 8 bit field

each field has 256 decimal equivalent. that is

binary                                        denary or decimal

11111111      =        2⁸      =             256

w.x.y.z represents

in binary

11111111.11111111.11111111.11111111

in denary

255.255.255.255

note that 255 = 2⁸ - 1 = no of valid hosts/addresses

there are classes of addresses, that is

class A = w.0.0.0 example 10.0.0.0

class B = w.x.0.0 example 172.16.0.0

class C = w.x.y.0 example 198.16.8.1

where w, x, y, z could take numbers from 1 to 255

Now in the question

we were given the ip address : 198.16.0.0 (class B)

address of quantity 4000, 2000, 8000 is possible with a subnet mask of type

255.255.0.0 (denary) or

11111111.11111111.00000000.00000000(binary) where /s =  /16 That is no of 1s

In a VLSM (Variable Length Subnet Mask)

Step 1

we convert the number of host/addresses for company A to binary

4000 = 111110100000 = 12 bit

step 2 (subnet mask)

vary the fixed subnet mask to reserve zeros (0s) for the 12 bit above

fixed subnet mask: 11111111.11111111.00000000.00000000            /16

variable subnet mask: 11111111.11111111.11110000.000000                /20

now we have added 4 1s in the 3rd field to reserve 12 0s

<u><em>subnet mask: 255.255.</em></u><u><em>16.</em></u><u><em>0 (where the 1s in each field represent a denary number as follows)</em></u>

<u><em>1st 1 = 128, 2nd 1 = 64 as follows</em></u>

<u><em>1        1       1      </em></u><u><em> 1 </em></u><u><em>      1     1     1     1</em></u>

<u><em>128  64     32    </em></u><u><em>16</em></u><u><em>    8    4     2    1</em></u>

step 3

in the ip network address: 198.16.0.0/19 <em>(subnet representation)</em> we increment this using 16

that is 16 is added to the 3rd field as follows

That means the ist Valid Ip address starts from

          Ist valid Ip add: 198.16.0.1 - 198.16.15.255(last valid IP address)

Company B starts<u><em>+16: 198.16.</em></u><u><em>16</em></u><u><em>.0 - 198.16.31.255</em></u>

<u><em>                   +16: 198.16.</em></u><u><em>32</em></u><u><em>.0- 198.16.47.255 et</em></u>c

we repeat the steps for other companies as follows

Company B

Step 1

we convert the number of host/addresses for company B to binary

2000 = 11111010000 = 11 bit

Step 2

vary the fixed subnet mask to reserve zeros (0s) for the 11 bit above

fixed subnet mask: 11111111.11111111.00000000.00000000            /16

variable subnet mask: 11111111.11111111.11111000.000000                /21

now we have added 5 1s in the third field to reserve 11 0s

<u><em>subnet mask: 255.255.</em></u><u><em>8.</em></u><u><em>0 (where the 1s in each field represent a denary number as follows)</em></u>

<u><em>1st 1 = 128, 2nd 1 = 64 as follows</em></u>

<u><em>1        1       1       1       </em></u><u><em>1 </em></u><u><em>    1     1     1</em></u>

<u><em>128  64     32    16    </em></u><u><em>8 </em></u><u><em>   4     2    1</em></u>

Step 3

Starting from after the last valid Ip address for company A

in the ip network address: 198.16.16.0/21 (<em>subnet representation</em>) we increment this using 8

That means the ist Valid Ip address starts from

           Ist valid Ip add: 198.16.16.1 - 198.16.23.255(last valid IP address)

Company C starts <u><em>+16: 198.16.</em></u><u><em>24</em></u><u><em>.0- 198.16.31.255</em></u>

<em>                             </em><u><em> +16: 198.16.</em></u><u><em>32</em></u><u><em>.0- 198.16.112.255 et</em></u>c

Company C

Step 1

we convert the number of host/addresses for company C to binary

4000 = 111110100000 = 12 bit

Step 2

vary the fixed subnet mask to reserve zeros (0s) for the 12 bit above

fixed subnet mask: 11111111.11111111.00000000.00000000            /16

variable subnet mask: 11111111.11111111.11110000.000000                /20

now we have added 4 1s in the 3rd field to reserve 12 0s

<u><em>subnet mask: 255.255.</em></u><u><em>16.</em></u><u><em>0 (where the 1s in each field represent a denary number as follows)</em></u>

<u><em>1st 1 = 128, 2nd 1 = 64 as follows</em></u>

<u><em>1        1       1       1       1     1     1     1</em></u>

<u><em>128  64     32    16    8    4     2    1</em></u>

Step 3

Starting from after the last valid ip address for company B

in the ip network address: 198.16.24.0/20 (subnet representation) we increment this using 16

That means the ist Valid Ip address starts from

           Ist valid Ip add: 198.16.24.1 - 198.16.39.255(last valid IP address)

Company C starts <u><em>+16: 198.16.40.0- 198.16.55.255</em></u>

<em>                          </em><u><em>    +16: 198.16.56.0- 198.16.71.255 et</em></u>c

Company D

Step 1

we convert the number of host/addresses for company D to binary

8000 = 1111101000000 = 13 bit

Step 2

vary the fixed subnet mask to reserve zeros (0s) for the 13 bit above

fixed subnet mask: 11111111.11111111.00000000.00000000            /16

variable subnet mask: 11111111.11111111.11100000.000000                /19

now we have added 3 1s in the 3rd field to reserve 13 0s

<u><em>subnet mask: 255.255.</em></u><u><em>32.</em></u><u><em>0 (where the 1s in each field represent a denary number as follows)</em></u>

<u><em>1st 1 = 128, 2nd 1 = 64 as follows</em></u>

<u><em>1        1      </em></u><u><em> 1 </em></u><u><em>      1       1     1     1     1</em></u>

<u><em>128  64     </em></u><u><em>32  </em></u><u><em>  16    8    4     2    1</em></u>

Step 3

Starting from after the last valid ip address for company C

in the ip network address: 198.16.40.0/20 (subnet representation) we increment this using 32

That means the ist Valid Ip address starts from

           Ist valid Ip add: 198.16.40.1 - 198.16.71.255(last valid IP address)

Company C starts <u><em>+16: 198.16.72.0- 198.16.103.255</em></u>

<em>                          </em><u><em>    +16: 198.16.104.0- 198.16.136.255 et</em></u>c

5 0
3 years ago
)Dynamically allocate an object of Banana, using the pointer variable daco.
zepelin [54]

Answer:

d) daco = new Banana;

Explanation:

Dynamically allocated variables have their memory allocated in the heap memory.We declare a dynamical variable like this:-

int *a=new int ;

It means a pointer a is created on the stack memory which hold the address of the block that hold the value of variable a in heap memory.

We already have the pointer daco. We just have to initialize with keyword new.

It will be like daco=new Banana; which matches the option d.

5 0
3 years ago
Each symbol of an octal number corresponds to 3 bits of a binary number. is it true or false​
grin007 [14]

Answer: True.

Explanation: It uses only the 3 bits to represent any digit in binary and easy to convert from octal to binary and then to vice-versa. Hope that helps

5 0
3 years ago
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