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never [62]
2 years ago
9

An analyst receives an alert from the SIEM showing an IP address that does not belong to the assigned network can be seen sendin

g packets to the wrong gateway. Which of the following network devices is misconfigured and which of the following should be done to remediate the issue? A. Firewall; implement an ACL on the interface B. Router; place the correct subnet on the interface C. Switch; modify the access port to trunk port D. Proxy; add the correct transparent interface
Computers and Technology
1 answer:
lianna [129]2 years ago
4 0

Answer: (A)  Firewall; implement an ACL on the interface

Explanation:

 According to the question, for re-mediate the issue firewall should be implemented the ACL (Access control list) on the given interface. The access control list is one of the type of logic which selectively give permission or deny the number of packet in the system which to through the interface.

Firewall is the type of device that basically examine the traffic in the network system and also make several decisions in the system. ACL is the set of rule which mainly define the route of the packet in the router interface state.

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Ryan has made a presentation of ten slides, which he wants to display in sequence after a specific time interval. Which element
alukav5142 [94]

Answer:

Slide transition.

Explanation:

PowerPoint application can be defined as a software application or program designed and developed by Microsoft, to avail users the ability to create various slides containing textual and multimedia informations that can be used during a presentation.

Some of the features available on Microsoft PowerPoint are narrations, transition effects, custom slideshows, animation effects, formatting options etc.

Basically, the views that are available on the Microsoft PowerPoint application includes;

1. Slide Sorter.

2. Notes Page.

3. Reading Pane.

4. Presenter view.

In this scenario, Ryan has made a presentation of ten slides, which he wants to display in sequence after a specific time interval.

Using slide transition in the presentation software, Ryan can make the slides appear, one after the other, after a specific time interval.

Slide transition is an inbuilt feature of a presentation software that automatically changes the slides at regular intervals.

3 0
2 years ago
1. Write a set of routines for implementing several stacks and queues within a single array. Hint: Look at the lecture material
JulsSmile [24]

Answer:

Out of the two methods The 2nd method is a better option for implementing several stacks and arrays using the single array system this is because Method 2 makes efficient use of the available space

Explanation:

A) Method 1 ( Divide the given single array in the size of n/k.)

  • How to use method 1 to implement several stacks involves :

i)  To implement several stacks through an array (x) is by dividing the array in n/k parts.

ii) The k represents the slots in which the different stacks will be placed and the n represents the size of the array x.

iii) If you need to implement at least two stacks place the first stack in the slot of a [0] to a [n/k - 1], and another stack in the slot of a[n/k] to a[2n/k-1].

note: The  disadvantage of this method is that the use of the space of the array is not much efficient. therefore method 2 is better

  • How to use method 1 to implement queues involves :

i) queues can also be implemented through an array. applying the same method above

ii) Divide the array in slots and place the queues in that slots.

This method has a problem with  the efficient utilization of the space.therefore method 2 is preferred

B) Method 2 ( uses the space efficiently ) uses two more arrays to implement stacks which are : Top_array and Next_array

  • how to use method 2 to implement several stacks

i) Store the indexes of the next item that will also be stored in all stacks in this initial stack

ii)The initial actual array is x[] and this will store the stacks.

iii)Simultaneously with several stacks, the stack which contain the free slots in the array x[] will also be maintained.

iv)  The entries of the Top_array[] will be initialized to -1. This implies that all the stacks are empty.

v)  Firstly the entries of the array Next_array[i] will be initialized to i+1, since all the slots initially are free and are pointed to the next slot.

vi) Initialize The top of the free stack that is maintaining the free slots  as 0.

vii)  The complexity of push () (method to insert an element) and pop () (method to delete an element) operations by using this method is O (1).

  • How to use method 2 to implement several queues

The same method applicable to implementing several stacks is used here but with a difference is the presence of three extra arrays which are :

Front_array[] = indicates the number of queues. This array stores the indexes of the front elements of the stacks.

Rear_array[] = determines the sizeof the array k . This array stores the indexes of the last elements of the stacks.

Next_array[] = The array n indicates the size of the single array say x. This array stores the indexes of the next items that is being pushed.

i) The initial actual array is a[] which will store the queues. The free slots will also be maintained.

ii)The entries of the Front_array[] will be initialized to -1. This means  that all the queues are empty initially

ii) Initially the entries of the array Next_array[i] will be initialized to i+1, since all the slots initially are free and are pointed to the next slot.

iv) Apply The complexity of enqueue ()  and dequeue ()  by using this method is O (1).

4 0
2 years ago
When creating a table in the relational database design from an entity in the extended e-r model, the attributes of the entity b
larisa86 [58]
No.  The attributes of the table correspond to columns within the table.  Each unique set of attribute values taken together correspond to table rows.  Sometimes referred to as "records".

8 0
2 years ago
Write a program that simulates flipping a coin to make decisions. The input is how many decisions are needed, and the output is
Nataly [62]

Answer:

import random

decisions = int(input("How many decisions: "))

for i in range(decisions):

   number = random.randint(0, 1)

   if number == 0:

       print("heads")

   else:

       print("tails")

Explanation:

*The code is in Python.

import the random to be able to generate random numbers

Ask the user to enter the number of decisions

Create a for loop that iterates number of decisions times. For each round; generate a number between 0 and 1 using the randint() method. Check the number. If it is equal to 0, print "heads". Otherwise, print "tails"

7 0
2 years 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
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