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Nastasia [14]
3 years ago
5

Each of the following is a benefit provided by using views except for one. Which one is it? Group of answer choices a) You can u

se views to restrict access to the tables in your database to certain rows. b) You can code views that join tables. c) Views can be used to hide the complexity of retrieval operations. d) Views can be used to update multiple tables with a single statement.
Computers and Technology
1 answer:
Dmitry_Shevchenko [17]3 years ago
8 0

Answer:

The answer to this question can be given as:

In this question the option b,c, and d are correct.

Explanation:

A (DBMS) stands for a database management system. The DBMS is a software package that is used for crate, update, and delete. The database uses the SQL which stands for the structured query language. It is a programming language that uses the queries. These queries are used to providing many operations in the database like crate table, update table, join table, create a view of the table, etc. In the database, we can view the code of the join table and is also used to hide the complexity of retrieval operations, it can help in the update the table in a single statement.

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Explanation:

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3 years ago
Write IF() function for student 3, If total mark is greater than 10 then print
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Answer:

Explanation:

The following code is written in Java. The if() function is actually called an if statement. The following code can be copied and pasted where you need as long as it is within the scope of the mark variable in order to call it and compare it to 10

if (mark > 10){

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3 years ago
Swapping two numbers
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Answer:

Swapping two numbers means exchange the values of two variables with each other.

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3 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

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A Microsoft Office Suite.

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