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PolarNik [594]
3 years ago
12

When you send an email how many computers does it go to

Computers and Technology
1 answer:
Xelga [282]3 years ago
4 0

once you have sent the email it will only go to the server which then the person receiving it would log on therefore requesting the emails recived.

I'm not 100% sure my answer is classed as correct but I would say it doesn't go to a computer it just gets stored on a server in till requested

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Suppose there are 69 packets entering a queue at the same time. Each packet is of size 7 MiB. The link transmission rate is 1.7
svp [43]

Answer:

69.08265412 milliseconds

Explanation:

Lets first convert 7 MiB to bits

7*1024*1024*8=58720256 bits

Now convert bits to Gbits

58720256/10^{9}  =0.058720256 Gbits

Queuing Delay = Total size/transmission link rate

Queuing Delay= \frac{0.058720256}{1.7} =0.03454132706 seconds

Delay of packet number 3 = 0.03454132706*2=0.06908265412 seconds

or 0.06908265412= 69.08265412 milliseconds

7 0
3 years ago
I NEED HELP!!! BRAINLIEST!!!
Trava [24]

Answer:

DHCP Dynamic Host Configuration Protocol:

Is a network service that automatically assigns IP addresses and other TCP/IP configuration information on network nodes configured as DHCP clients. Server allocates IP addresses to DHCP clients dynamically. Should be configured with at least one DHCP scope. Scope contains a range of IP addresses and a subnet mask, and can contain other options, such as a default gateway and Domain Name System. Scope also needs to specify the duration of the lease and usage of an IP affects after which the node needs to renew the lease with the SHCP server. Determines the duration, which can be set for a defined time period or for an unlimited length of time.

DNS Domain Name Service: Is a TCP/IP name resolution service that translates FQDNs into IP addresses. System of hierarchical databases that are stored on separate DNS servers on all networks that connect to the Internet. DNS servers store, maintains and update databases, they respond to DNS client name resolution requests to translate host names into IP addresses.

DNS Components

DNS database is divided logically into a heieratchical grouping of domains. Physically into files called zones. Zone files contain the actual IP-to-host name mapping for one or more domains. Zone files is stored on the DNS server that is responsible for resolving hot names for the domains contained in the zone. Each network node in that domain will have a host record within the domain's zone files. Includes the node's host name, FQDN, and assigned IP address.

DNS Servers

*If you are configuring static IP addresses, including the IP address of the default DNS servers as you configure each client.

*If you are using DHCP, use the DHCP scope options to specify the IP Explanation:

3 0
2 years ago
Read 2 more answers
It's the same drop-down answers for both.
tia_tia [17]

Answer:

First, medium: can see all the subject but maintain the focus in him

Second, close-up: To se the facial expressions and emotions

5 0
3 years ago
What are the main techniques used to help manage test anxiety? Check all that apply.
Aleks04 [339]

Answer:

1) Avoid the perfectionist of trap

2)Banish the negative thoughts

3) Get enough sleep

4) Make sure you are prepared

5) take deep breaths

Explanation:

In my thought it's the answer of this question.

4 0
2 years ago
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Instructions The population of town A is less than the population of town B. However, the population of town A is growing faster
weeeeeb [17]

Answer: The c++ program is given below.

#include <iostream>

using namespace std;

int main() {

float townA, townB, growthA, growthB, populationA, populationB;

    int year=0;    

cout<<"Enter present population of town A : ";

cin >> townA;  

cout<<endl<<"Enter present growth rate of town A : ";

cin >> growthA;

growthA = growthA/100;  

cout<<endl<<"Enter present population of town B : ";

cin >> townB;  

cout<<endl<<"Enter present growth rate of town B : ";

cin >> growthB;

growthB = growthB/100;  

do

{

    populationA = townA + (townA * growthA);

    populationB = townB + (townB * growthB);      

    townA = populationA;

    townB = populationB;      

    year++;      

}while(populationA < populationB);  

cout<<endl<<"After " <<year<< " years, population of town A is "<<populationA << " and population of town B is "<< population<<endl;

return 0;

}

Explanation:

All the variables for population and growth rate are declared with float datatype.

The user inputs the present population of both the towns.

The growth rate entered by the user is the percentage growth.

For example, town A has 10% growth rate as shown in the output image.

The program converts 10% into float as 10/100.

growthA = growthA/100;

growthB = growthB/100;

The above conversion is done to ease the calculations.

The year variable is declared as integer and initialized to 0.

The growth in population is computed in a do-while loop. After each growth is calculated, the year variable is incremented by 1.

The loop continues until population of town A becomes greater than or equal to population of town B as mentioned in the question.

Once the loop discontinues, the final populations of town A and town B and the years needed for this growth is displayed.

The new line is introduced using endl keyword.

The function main has return type int hence, 0 is returned at the end of the program.

6 0
3 years ago
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