Please help me please it's urgent

Which of the following examines the organizational resource of information and regulates its definitions, uses, value, and distr

hichkok12 [17]

Answer:C)Information management

Explanation: Information management is the technique through which the organization and handling of the the data/ information takes place.This technique is made for helping the organization and their business for the supporting the function and processes.

The facilities provided by the information management system is development, management, designing, innovation etc.

Other options are incorrect because they are used for the coding process,technology and governing of the data, hence no management activity is done by them.Thus, the correct option is option(C).

4 0

3 years ago

Print numbers 0, 1, 2, ..., userNum as shown, with each number indented by that number of spaces. For each printed line, print t

geniusboy [140]

Answer:

Please find the answer below

Explanation:

// Online C compiler to run C program online

#include <stdio.h>

int main() {

// Write C code here

//printf("Hello world");

int userNum;

int i;

int j;

scanf("%d", &userNum);

/* Your solution goes here */

for(i = 0; i<=userNum; i++){

for(j = 0; j <= i; j++){

printf(" ");

}

printf("%d\n", i);

}

return 0;

}

8 0

2 years ago

3. Which of the following is a single piece of information related to the person, place,

stepan [7]

I think the answer is field

8 0

3 years ago

Read 2 more answers

Is the wireless network you own as secure as it should be? Examine your wireless network or that of a friend or neighbor and det

AleksandrR [38]

Answer:

See explaination

Explanation:

Wireless Security:

In a bid to examine the network you can use ipconfig /all command in windows or in the event where you are a linux user type in ifconfig to view all the configuration details of your network such as

indows IP Configuration

Host Name . . . . . . . . . . . . : myvpc-hb

Primary Dns Suffix . . . . . . . : example.com

Node Type . . . . . . . . . . . . : Hybrid

IP Routing Enabled. . . . . . . : Yes

Ethernet adapter Local Area Connection:- - - - - -

Connection-specific DNS Suffix . :- - - - - -

Description . . . . . . . . . . : Intel 68540-Based PCI Express Fast Ethernet Adapter

Physical Address. . . . . . . . . : 00-03-AA-BC-CA-8F

Autoconfiguration Enabled . . . . : Yes

Link-local IPv6 Address . . . . . : fe65::ab56:ccb9:320c:524d%8(Preferred)

IPv4 Address. . . . . . . . . . . : 192.168.45.111(Preferred)

Default Gateway . . . . . . . . . : 192.168.15.1

DHCP Server . . . . . . . . . . . : 192.168.15.1

DHCPv6 IAID . . . . . . . . . . . : 201327615

Note: If you want to deeply examine the your network and to find security model go with Wireshark application.It will help you to capture the data packets easily using GUI.Also it will provide other details like transmission control protocols,etc.

To take your network to highest level i.e. to make it more secure you can use:

1.Firewalls like application gateways,packet filtering,hybrid systems.

2.Crypto cable routers

3.Virtual private networks, etc.

Estimation of cost and the time to increase the level is solely depends on the type of the architecture you want to use for the network building structure.Also the size of the network will be responsible for the cost and time.

In case of any attack on your computer it depends on the size of the data on the computer disc and the bandwidth of the network to which your computer is connected to replace all data. The faster the bandwidth of the network,faster the data replacement rate.

3 0

3 years ago

Computer Networks - Queues

lyudmila [28]

Answer:

the average arrival rate \lambda in units of packets/second is 15.24 kbps

the average number of packets w waiting to be serviced in the buffer is 762 bits

Explanation:

Given that:

A single channel with a capacity of 64 kbps.

Average packet waiting time $T_w$ in the buffer = 0.05 second

Average number of packets in residence = 1 packet

Average packet length r = 1000 bits

What are the average arrival rate \lambda in units of packets/second and the average number of packets w waiting to be serviced in the buffer?

The Conservation of Time and Messages ;

E(R) = E(W) + ρ

r = w + ρ

Using Little law ;

r = λ × T_r

w = λ × T_w

r / λ = w / λ + ρ / λ

T_r =T_w + 1 / μ

T_r = T_w +T_s

where ;

ρ = utilisation fraction of time facility

r = mean number of item in the system waiting to be served

w = mean number of packet waiting to be served

λ = mean number of arrival per second

T_r =mean time an item spent in the system

T_w = mean waiting time

μ = traffic intensity

T_s = mean service time for each arrival

the average arrival rate \lambda in units of packets/second; we have the following.

First let's determine the serving time T_s

the serving time T_s $= \dfrac{1000}{64*1000}$

= 0.015625

now; the mean time an item spent in the system T_r = T_w +T_s

where;

T_w = 0.05 (i.e the average packet waiting time)

T_s = 0.015625

T_r = 0.05 + 0.015625

T_r = 0.065625

However; the mean number of arrival per second λ is;

r = λ × T_r

λ = r / T_r

λ = 1000 / 0.065625

λ = 15238.09524 bps

λ ≅ 15.24 kbps

Thus; the average arrival rate \lambda in units of packets/second is 15.24 kbps

b) Determine the average number of packets w waiting to be serviced in the buffer.

mean number of packets w waiting to be served is calculated using the formula

w = λ × T_w

where;

T_w = 0.05

w = 15238.09524 × 0.05

w = 761.904762

w ≅ 762 bits

Thus; the average number of packets w waiting to be serviced in the buffer is 762 bits

4 0

3 years ago

Please help me please it's urgent​

Please help me please it's urgent