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shtirl [24]
1 year ago
11

When trying to improve performance of a slow system, you notice in Task Manager that the superfetch service is using a high perc

entage of CPU time. What is your next best step?
a. Disable superfetch to improve performance.

b. Update Windows to improve superfetch performance.

c. Superfetch is an essential Windows process and should not be disabled. Move on to other solutions to improve performance.

d. Ask the user if he uses the superfetch service. If he doesn’t, uninstall it.
Computers and Technology
1 answer:
ehidna [41]1 year ago
4 0

Answer:

Option A i.e.,Disable the superfetch to improve performance is the correct option.

Explanation:

The following option is correct because when the user is trying to maintain the performance of their system then, he observe that the following tool that is Task Manager display that the superfetch using the CPU time in high rate then, he decided to disable the  Window service tool that is superfetch to increase the responding time of the system for improving the performance.

You might be interested in
The Company management has asked that you compare the OSSTMM and the PTES to determine which methodology to select for internal
Sonbull [250]

Answer:

The basic comaprism of OSSTMN and PTES includes the following: OSSTMN is more theoretical, security assessment methodology, and Metrics based why PTES is technology oriented, penetration testing methodology ,  extended analysis of all stages

Explanation:

Solution

Penetration testing has several methodologies which include :OSSTMM and PTES  

The comparison between OSSTMM and PTES is stated as follows:

OSSTMM:                                                

Security assessment methodology

More Theoretical  

Metrics based

PTES :

Technology oriented

Penetration testing methodology

Extended analysis of all stages

Now,

There are 7 stages which is used to define PTES for penetration testing.(Penetration Testing Execution Standard)

  • Pre-engagement Interactions
  • Intelligence Gathering
  • Threat Modeling
  • Vulnerability Analysis
  • Exploitation
  • Post Exploitation
  • Reporting

Now,

The OSSTMM is used to obtain security metrics and performing penetration testing .The OSSTMM provides transparency to those who have inadequate security policies and configurations.

The OSSTMM includes the entire risk assessment process starting from requirement analysis to report creation.

Six areas are covered by OSSTMM which are:

  • Information security
  • Process security
  • Internet technology security
  • Communications security
  • Wireless security
  • Physical security
7 0
1 year ago
8.Change the following IP addresses from binary notation to dotted-decimal notation: a.01111111 11110000 01100111 01111101 b.101
Andrews [41]

Answer:

a. 01111111 11110000 01100111 01111101 dotted decimal notation:

(127.240.103.125)

b. 10101111 11000000 11111000 00011101 dotted decimal notation: (175.192.248.29)

c. 11011111 10110000 00011111 01011101 dotted decimal notation:

(223.176.31.93)

d. 11101111 11110111 11000111 00011101 dotted decimal notation:

(239.247.199.29)

a. 208.34.54.12 class is C

b. 238.34.2.1 class is D

c. 242.34.2.8 class is E

d. 129.14.6.8 class is B

a.11110111 11110011 10000111 11011101 class is E

b.10101111 11000000 11110000 00011101 class is B

c.11011111 10110000 00011111 01011101 class is C

d.11101111 11110111 11000111 00011101 class is D

Explanation:

8 a. 01111111 11110000 01100111 01111101

we have to convert this binary notation to dotted decimal notation.

01111111 = 0*2^7 + 1*2^6 + 1*2^5 + 1*2^4 + 1*2^3 + 1*2^2 + 1*2^1 + 1*2^0

           = 0 + 1*64 + 1*32 + 1*16 + 1*8 + 1*4 + 1*2 + 1

           = 64 + 32 + 16 + 8 + 4 + 2 + 1

           = 127

11110000 = 1*2^7 + 1*2^6 + 1*2^5 + 1*2^4 + 0*2^3 + 0*2^2 + 0*2^1 + 0*2^0

               = 1*128 + 1*64 + 1*32 + 16 + 0 + 0 + 0 + 0

               = 128 + 64 + 32 + 16

               = 240

01100111 = 0*2^7 + 1*2^6 + 1*2^5 + 0*2^4 +0*2^3 + 1*2^2 + 1*2^1 + 1*2^0

               = 0 + 1*64 + 1*32 + 0 + 0 + 4 + 2 + 1

               = 64 + 32 + 4 + 2 + 1

               = 103

01111101   = 0*2^7 + 1*2^6 + 1*2^5+ 1*2^4 +1*2^3 +1*2^2 + 0*2^1 + 1*2^0

               = 0 + 1*64 + 1*32 + 1*16 + 1*8 + 1* 4 + 0 + 1

               = 64 + 32 + 16 + 8 + 4 + 1

               = 125

So the IP address from binary notation 01111111 11110000 01100111 01111101 to dotted decimal notation is : 127.240.103.125

b) 10101111 11000000 11111000 00011101

10101111 = 1*2^7 + 0*2^6 + 1*2^5 + 0*2^4 + 1*2^3 + 1*2^2 + 1*2^1 + 1*2^0

             = 175

11000000 = 1*2^7 + 1*2^6 + 0*2^5 + 0*2^4 + 0*2^3 + 0*2^2 + 0*2^1 + 0*2^0

                 = 192

11111000 = 1*2^7 + 1*2^6 + 1*2^5 + 1*2^4 +1*2^3 + 0*2^2 + 0*2^1 + 0*2^0

              = 248

00011101 = 0*2^7 + 0*2^6 + 0*2^5 + 1*2^4 +1*2^3 + 1*2^2 + 0*2^1 + 1*2^0

               = 29

So the IP address from binary notation 10101111 11000000 11111000 00011101  to dotted decimal notation is : 175.192.248.29

c) 11011111 10110000 00011111 01011101

11011111 = 1*2^7 + 1*2^6 + 0*2^5 + 1*2^4 +1*2^3 + 1*2^2 + 1*2^1 + 1*2^0

           = 223

10110000 =  1*2^7 + 0*2^6 + 1*2^5 + 1*2^4 +0*2^3 + 0*2^2 + 0*2^1 + 0*2^0

                = 176

00011111 = 0*2^7 + 0*2^6 + 0*2^5 + 1*2^4 +1*2^3 + 1*2^2 + 1*2^1 + 1*2^0

              = 31

01011101 = 0*2^7 + 1*2^6 + 0*2^5 + 1*2^4 +1*2^3 + 1*2^2 + 0*2^1 + 1*2^0

              = 93

So the IP address from binary notation 11011111 10110000 00011111 01011101 to dotted decimal notation is :223.176.31.93

d) 11101111 11110111 11000111 00011101

11101111 = 1*2^7 + 1*2^6 + 1*2^5 + 0*2^4 +1*2^3 + 1*2^2 + 1*2^1 + 1*2^0

            = 239

11110111 = 1*2^7 + 1*2^6 + 1*2^5 + 1*2^4 +0*2^3 + 1*2^2 + 1*2^1 + 1*2^0

           = 247

11000111 =  1*2^7 + 1*2^6 + 0*2^5 + 0*2^4 +0*2^3 + 1*2^2 + 1*2^1 + 1*2^0

              = 199

00011101 = 0*2^7 + 0*2^6 + 0*2^5 + 1*2^4 +1*2^3 + 1*2^2 + 0*2^1 + 1*2^0

               = 29

So the IP address from binary notation 11101111 11110111 11000111 00011101 to dotted decimal notation is : 239.247.199.29

9. In order to the find the class check the first byte of the IP address which is first 8 bits and check the corresponding class as follows:                

Class A is from 0 to 127

Class B is from 128 to 191

Class C is from 192 to 223

Class D is from 224 to 239

Class E is from 240 to 255

a. 208.34.54.12

If we see the first byte of the IP address which is 208, it belongs to class C as class C ranges from 192 to 223.

b. 238.34.2.1

If we see the first byte of the IP address which is 238, it belongs to class D as Class D ranges from 224 to 239.

c. 242.34.2.8

If we see the first byte of the IP address which is 242, it belongs to class E as Class E ranges from 240 to 255.

d. 129.14.6.8

If we see the first byte of the IP address which is 129, it belongs to class B as Class B ranges from 128 to 191.

10. In order to find the class of the IP addresses in easy way, start checking bit my bit from the left of the IP address and follow this pattern:

0 = Class A

1 - 0 = Class B

1 - 1 - 0 = Class C

1 - 1 - 1 - 0 = Class D

1 - 1 - 1 - 1 = Class E

a. 11110111 11110011 10000111 11011101

If we see the first four bits of the IP address they are 1111 which matches the pattern of class E given above. So this IP address belongs to class E.

b. 10101111 11000000 11110000 00011101

If we see the first bit is 1, the second bit is 0 which shows that this is class B address as 1 0 = Class B given above.

c. 11011111 10110000 00011111 01011101

The first bit is 1, second bit is 1 and third bit is 0 which shows this address belongs to class C as 110 = Class C given above.

d. 11101111 11110111 11000111 00011101

The first bit is 1, the second bit is also 1 and third bit is also 1 which shows that this address belongs to class D.

3 0
2 years ago
Type the correct answer in the box. Spell all words correctly.
wlad13 [49]

Ryan should apply a filter/criteria on the Courses column and view all the courses that show Sociology

3 0
1 year ago
Write the 8-bit signed-magnitude, two's complement, and ones' complement representations for each decimal number: +25, + 120, +
pshichka [43]

Answer:

Let's convert the decimals into signed 8-bit binary numbers.

As we need to find the 8-bit magnitude, so write the powers at each bit.

      <u>Sign -bit</u> <u>64</u> <u>32</u> <u>16</u> <u>8</u> <u>4</u> <u>2</u> <u>1</u>

+25 - 0 0 0 1 1 0 0 1

+120- 0 1 1 1 1 0 0 0

+82 - 0 1 0 1 0 0 1       0

-42 - 1 0 1 0 1 0 1 0

-111 - 1 1 1 0 1 1 1 1

One’s Complements:  

+25 (00011001) – 11100110

+120(01111000) - 10000111

+82(01010010) - 10101101

-42(10101010) - 01010101

-111(11101111)- 00010000

Two’s Complements:  

+25 (00011001) – 11100110+1 = 11100111

+120(01111000) – 10000111+1 = 10001000

+82(01010010) – 10101101+1= 10101110

-42(10101010) – 01010101+1= 01010110

-111(11101111)- 00010000+1= 00010001

Explanation:

To find the 8-bit signed magnitude follow this process:

For +120

  • put 0 at Sign-bit as there is plus sign before 120.
  • Put 1 at the largest power of 2 near to 120 and less than 120, so put 1 at 64.
  • Subtract 64 from 120, i.e. 120-64 = 56.
  • Then put 1 at 32, as it is the nearest power of 2 of 56. Then 56-32=24.
  • Then put 1 at 16 and 24-16 = 8.
  • Now put 1 at 8. 8-8 = 0, so put 0 at all rest places.

To find one’s complement of a number 00011001, find 11111111 – 00011001 or put 0 in place each 1 and 1 in place of each 0., i.e., 11100110.

Now to find Two’s complement of a number, just do binary addition of the number with 1.

6 0
2 years ago
A(n) ____________________ is a named collection of stored data, instructions, or information and can contain text, images, video
Vanyuwa [196]
The answer in the blank is files for they exist in the computer that are responsible to hold or collect different types of application that could contribute in arranging them in order or showing classifications or simply storing them to put them in one place. This is essential to the computer for they are responsible in putting them in place and showing organization.
7 0
2 years ago
Read 2 more answers
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