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adell [148]
3 years ago
11

Fill in the blank. The process that a wireless router uses to translate a private IP address on internal traffic to a routable a

ddress for the Internet is called?
Computers and Technology
1 answer:
AleksandrR [38]3 years ago
6 0

Answer:

Network Address Translation (NAT)

Explanation:

With the aid of Network address translation(NAT), the IP addresses of a particular local network are translated or mapped into a single or multiple global or public IP addresses. Therefore, a wireless router could use NAT to translate its private IP address on internal traffic (network) to a routable address for the internet.

With NAT, businesses can use many internal IP addresses since they are just for internal purposes and will be eventually converted into a single or a few multiple routable IP addresses.

Three types of NAT are possible:

(i) Static NAT : There is a one - to - one mapping between private IP addresses and routable (public) IP addresses. One private IP is mapped to one public IP address.

(ii) Dynamic NAT : There is a many- to - many mapping between private IP addresses and routable (public) IP addresses. Multiple private IPs are mapped to many public IP addresses.

(iii) Port Address Translation (PAT) : Many - to - one relationship between the private IP addresses and public addresses. Many private IP addresses can be mapped or translated into a single public IP address. This type of NAT is also called NAT overload.

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What is the limitation of computer<br>​
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The limitation of computer are:

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3 years ago
Perform online or offline research to learn more about film elements. Then, write a short report on the essentials of filmmaking
pshichka [43]

Answer:

Screenplay

Financing

Theme

cinematography

editing

sound and music

Acting and Genre

Explanation:

There are various essential elements that needs to be considered to become a famous and a good filmmaker. A good filmmaker will take care of small elements and will consider minor details. The audience is very keen when it takes interest in the movie. They pay attention to even minor details in the movie. The filmmaker should take care of all the details and will have to decide on the main things such as screenplay, audio, editing, genre of the movie and similar. These elements are dependent on each other. If the audio of the movie is not appropriate or sound quality is not suitable then the overall impact of the movie will be slowed no matter how strong the content is. There is high dependency of all the elements and every element should be paid attention.

3 0
3 years ago
Annie needs to provide a form field that will allow users to select from a predefined list of options but without the ability to
Lana71 [14]

Answer:

Drop down list

Explanation:

You have not listed what the possible answers are, however a drop down list would be an option to accomplish this.

8 0
2 years ago
1. How many bits would you need to address a 2M × 32 memory if:
Dominik [7]

Answer:

  1. a) 23       b) 21
  2. a) 43        b) 42
  3. a) 0          b) 0

Explanation:

<u>1) How many bits is needed to address a 2M * 32 memory </u>

2M = 2^1*2^20, while item =32 bit long word

hence ; L = 2^21 ; w = 32

a) when the memory is byte addressable

w = 8;  L = ( 2M * 32 ) / 8 =  2M * 4

hence number of bits =  log2(2M * 4)= log2 ( 2 * 2^20 * 2^2 ) = 23 bits

b) when the memory is word addressable

W = 32 ; L = ( 2M * 32 )/ 32 = 2M

hence the number of bits = log2 ( 2M ) = Log2 (2 * 2^20 ) = 21 bits

<u>2) How many bits are required to address a 4M × 16 main memory</u>

4M = 4^1*4^20 while item = 16 bit long word

hence L ( length ) = 4^21 ; w = 16

a) when the memory is byte addressable

w = 8 ; L = ( 4M * 16 ) / 8 = 4M * 2

hence number of bits = log 2 ( 4M * 2 ) = log 2 ( 4^1*4^20*2^1 ) ≈ 43 bits

b) when the memory is word addressable

w = 16 ; L = ( 4M * 16 ) / 16 = 4M

hence number of bits = log 2 ( 4M ) = log2 ( 4^1*4^20 ) ≈ 42 bits

<u>3) How many bits are required to address a 1M * 8 main memory </u>

1M = 1^1 * 1^20 ,  item = 8

L = 1^21 ; w = 8

a) when the memory is byte addressable

w = 8 ; L = ( 1 M * 8 ) / 8 = 1M

hence number of bits = log 2 ( 1M ) = log2 ( 1^1 * 1^20 ) = 0 bit

b) when memory is word addressable

w = 8 ; L = ( 1 M * 8 ) / 8 = 1M

number of bits = 0

5 0
2 years ago
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