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4 years ago

"is it important to have a firewall on a home network

1 answer:

5 0

Yes, you should have a firewall on your home network prevent hackers from getting in your network and causing damage to your devices or files. You should be able to put up a firewall with your Windows computer on the home menu>control panel>Firewall and that should work.

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You manage a network that uses a single switch. All ports within your building connect through the single switch. In the lobby o

Karo-lina-s [1.5K]

Answer:

VLANs

Explanation:

Virtual LANs (VLANs) are a type of network connection or arrangement of network devices in the same broadcast domain. They are called Virtual LANs because they represent fractions or subgroups in the switch ports found in an Ethernet LAN.

Indeed, because of their topological arrangement, this reduces the security risk of unauthorized access to sensitive data or devices since the host can be placed on a different VLAN. Hence, in this scenario, using this network feature would allow visitors to plug into these ports to gain internet access, but they would not have access to any other devices on the private network.

5 0

3 years ago

Each time the enter key is pressed, word creates a new paragraph. (points : 2) true false

ludmilkaskok [199]

Your answer would be true

6 0

3 years ago

The while loop and the do loop are equivalent in their expressive power; in other words, you can rewrite a while loop using a do

sleet_krkn [62]

I think False...

I hope this helps

8 0

2 years ago

1. How many bits would you need to address a 2M × 32 memory if:

Dominik [7]

Answer:

a) 23 b) 21
a) 43 b) 42
a) 0 b) 0

Explanation:

1) How many bits is needed to address a 2M * 32 memory

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

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

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

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

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

3 years ago

Mason is part of a project team that is creating a television advertisement. List one risk the team faces and a strategy for min

Andru [333]

One risk might be that they might use copyright music and and the music creator might take legal action.

5 0

3 years ago