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Leya [2.2K]
3 years ago
9

Access-lists pose a logical problem which often has more than one solution. Can you think of a different set of rules or placeme

nts that would yield the same required access filtering?
Computers and Technology
1 answer:
Maksim231197 [3]3 years ago
7 0

Answer:

Question is incomplete. it needs a topology diagram and also it needs Router R1 table. I assume User has access to the topology and Routing table.

Below Configuration will help to fix ACL problem

Hosts from the 172.16.0.0/16 network should have full access to Server1, Server2 and Server3 but this is not currently the case, as L1 can’t communicate to Server2 or Server3.

Explanation:

Following Configuration on Cisco Router R1 will help to fix all the ACL problems.

enable

configure terminal

no ip access-list standard FROM_10

ip access-list standard FROM_10

deny host 10.0.0.2

permit any

exit

!

no ip access-list standard FROM_172

ip access-list standard FROM_172

permit host 172.16.0.2

exit

!

interface GigabitEthernet0/0

ip access-group FROM_192 out

end

write memory

!

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Write a partial class that shows a class constant and an instance method. Write an instance method that converts feet to inches
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Answer:

Please the code snippet below, the code was writen in Kotlin Language

Explanation:

const val inches:Int= 12 .   //This is the const value

fun main(args: Array<String>) {

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   print("Enter a number")            

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var valueInFeet= Integer.valueOf(readLine())*inches

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2 years ago
Describe how you believe technology will impact our society in 50 years.
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Answer:

Wall-e

Explanation:

We will all be fat, lazy, on floating chairs with everything in our lives being automated thanks to AI and everyone's obsession to do less.

5 0
3 years ago
Write the function setKthDigit(n, k, d) that takes three integers -- n, k, and d -- where n is a possibly-negative int, k is a n
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Answer:

Explanation:

Let's do this in Python, first we need to convert the number into string in order to break it into a list of character, then we can replace the kth character with d. Finally we join all characters together, convert it to integer then output it

def setKthDigit(n, k, d):

n_string = str(n)

d_char = str(d)

n_char = [c for c in n_string]

n_char[k] = d_char

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6 0
3 years ago
We have said that the average number of comparisons need to find a target value in an n-element list using sequential search is
bija089 [108]

Answer:

Part a: If the list contains n elements (where n is odd) the middle term is at index (n-1)/2 and the number of comparisons are (n+1)/2.

Part b: If the list contains n elements (where n is even) the middle terms are  at index (n-2)/2 & n/2 and the number of comparisons are (n+2)/2.

Part c: The average number of comparisons for a list bearing n elements is 2n+3/4 comparisons.

Explanation:

Suppose the list is such that the starting index is 0.

Part a

If list has 15 elements, the middle item would be given at 7th index i.e.

there are 7 indices(0,1,2,3,4,5,6) below it and 7 indices(8,9,10,11,12,13,14) above it. It will have to run 8 comparisons  to find the middle term.

If list has 17 elements, the middle item would be given at 8th index i.e.

there are 8 indices(0,1,2,3,4,5,6,7) below it and 8 indices(9,10,11,12,13,14,15,16) above it.It will have to run 9 comparisons  to find the middle term.

If list has 21 elements, the middle item would be given at 10th index i.e.

there are 10 indices (0,1,2,3,4,5,6,7,8,9) below it and 10 indices (11,12,13,14,15,16,17,18,19,20) above it.It will have to run 11 comparisons  to find the middle term.

Now this indicates that if the list contains n elements (where n is odd) the middle term is at index (n-1)/2 and the number of comparisons are (n+1)/2.

Part b

If list has 16 elements, there are two middle terms as  one at would be at 7th index and the one at 8th index .There are 7 indices(0,1,2,3,4,5,6) below it and 7 indices(9,10,11,12,13,14,15) above it. It will have to run 9 comparisons  to find the middle terms.

If list has 18 elements, there are two middle terms as  one at would be at 8th index and the one at 9th index .There are 8 indices(0,1,2,3,4,5,6,7) below it and 8 indices(10,11,12,13,14,15,16,17) above it. It will have to run 10 comparisons  to find the middle terms.

If list has 20 elements, there are two middle terms as  one at would be at 9th index and the one at 10th index .There are 9 indices(0,1,2,3,4,5,6,7,8) below it and 9 indices(11,12,13,14,15,16,17,18,19) above it. It will have to run 11 comparisons  to find the middle terms.

Now this indicates that if the list contains n elements (where n is even) the middle terms are  at index (n-2)/2 & n/2 and the number of comparisons are (n+2)/2.

Part c

So the average number of comparisons is given as

((n+1)/2+(n+2)/2)/2=(2n+3)/4

So the average number of comparisons for a list bearing n elements is 2n+3/4 comparisons.

6 0
3 years ago
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