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lorasvet [3.4K]
3 years ago
8

Advantages and disadvantages of screen reading​

Computers and Technology
1 answer:
Vikki [24]3 years ago
8 0

Explanation:

do u mean screen reading as in reading books in ur mobile ?

if so then disadvantage is the amount of eye strain u get staring into the screen

and advantage is the portability.

You might be interested in
You are the IT administrator for a small corporate network. You have installed the Windows Server 2016 operating system on a ser
Elan Coil [88]

Answer:

In this lab, you perform the following tasks:

• On Disk 1, create a mirrored volume of the System (C:) volume to add fault tolerance.

• Using Disk 2, Disk 3, and Disk 4, create a RAID 5 volume that provides both fault tolerance and improved performance using the following settings:

o Volume size: 2 TB

o Drive letter: R

o Format: NTFS

o Volume label: Data

Complete this lab as follows:

1. Mirror an existing volume as follows:

a. Right-click Start and select Disk Management.

b. Click OK to initialize new disks.

c. Maximize the Disk Management window to better view the volumes.

d. Right-click the System (C:) volume and select Add Mirror.

e. Select Disk 1 that will be used for the mirrored copy.

f. Select Add Mirror.

g. Click Yes to convert the basic disk to a dynamic disk.

2. Create a RAID 5 volume as follows:

a. In Disk Management, right-click a disk with free space and select New RAID 5 Volume.

b. Click Next.

c. Under Available, holding down the Ctrl key, select Disk 3 and Disk 4 to be part of the new volume with Disk 2.

d. Select Add.

e. Click Next.

f. From the drive letter drop-down dialog, select R; then click Next.

g. Make sure that NTFS is selected as the file system.

h. In the Volume label field, enter Data.

i. Select Next.

j. Click Finish to create the volume.

k. Click Yes to convert the basic disk to a dynamic disk.

Explanation:

5 0
3 years ago
Given an array as follows
slava [35]

Answer:

1) Method calcTotal:

  1. public static long calcTotal(long [][] arr2D){
  2.        long total = 0;
  3.        
  4.        for(int i = 0; i < arr2D.length; i++ )
  5.        {
  6.            for(int j = 0; j < arr2D[i].length; j++)
  7.            {
  8.                total = total + arr2D[i][j];
  9.            }
  10.        }
  11.        
  12.        return total;
  13.    }

Explanation:

Line 1: Define a public method <em>calcTotal</em> and this method accept a two-dimensional array

Line 2: Declare a variable, total, and initialize it with zero.

Line 4: An outer for-loop to iterate through every rows of the two-dimensional array

Line 6: An inner  for-loop to iterate though every columns within a particular row.

Line 8: Within the inner for-loop, use current row and column index, i and j, to repeatedly extract the value of each element in the array and add it to the variable total.

Line 12: Return the final total of all the element values as an output

Answer:

2) Method calcAverage:

  1. public static double calcAverage(long [][] arr2D){
  2.        double total = 0;
  3.        int count = 0;
  4.        
  5.        for(int i = 0; i < arr2D.length; i++ )
  6.        {
  7.            for(int j = 0; j < arr2D[i].length; j++)
  8.            {
  9.                total = total + arr2D[i][j];
  10.                count++;
  11.            }
  12.            
  13.        }
  14.        
  15.        double average = total / count;
  16.        
  17.        return average;
  18.    }

Explanation:

The code in method <em>calcAverage</em> is quite similar to method <em>calcTotal</em>. We just need to add a counter and use that counter as a divisor of total values to obtain an average.

Line 4: Declare a variable, count, as an counter and initialize it to zero.

Line 11: Whenever an element of the 2D array is added to the total, the count is incremented by one. By doing so, we can get the total number of elements that exist in the array.

Line 16: Use the count as a divisor to the total to get average

Line 18: Return the average of all the values in the array as an output.

Answer:

3) calcRowAverage:

  1. public static double calcRowAverage(long [][] arr2D, int row){
  2.        double total = 0;
  3.        int count = 0;
  4.        
  5.        for(int i = 0; i < arr2D.length; i++ )
  6.        {
  7.            if(i == row)
  8.            {
  9.                for(int j = 0; j < arr2D[i].length; j++)
  10.                {
  11.                    total = total + arr2D[i][j];
  12.                    count++;
  13.                }
  14.            }
  15.            
  16.        }
  17.        
  18.        double average = total / count;
  19.        
  20.        return average;
  21.    }

Explanation:

By using method <em>calcAverage </em>as a foundation, add one more parameter, row, in the method <em>calcRowAverage</em>. The row number is used as an conditional checking criteria to ensure only that particular row of elements will be summed up and divided by the counter to get an average of that row.

Line 1: Add one more parameter, row,

Line 8-15: Check if current row index, i, is equal to the target row number, proceed to sum up the array element in that particular row and increment the counter.

5 0
3 years ago
Which area located at the top of the word screen includes home , insert , and page layout ?
Misha Larkins [42]

Answer:

Whats your qeustion

Explanation:

4 0
3 years ago
Read 2 more answers
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
Nina visited an area that receives a large amount of precipitation during all twelve months and is typically warm year round. Th
slava [35]

Answer: Rainforest

Explanation:

A rainforest is an area that typically has tall, green trees and also has a high amount of rainfall. Rainforest can be found in the wet tropical uplands and lowlands. The largest rainforests can be seen in the Amazon River Basin.

Since the area visited by Nina has a large amount of precipitation during all twelve months, typically warm year round and has a large variety of organism diversity, then the biome thatcNina most likely visited is the rainforest.

7 0
3 years ago
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