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Korvikt [17]
3 years ago
5

Funny or fascinating web content is said to go _________ when it is shared and re-shared a large number of times over a short pe

riod.
Computers and Technology
1 answer:
beks73 [17]3 years ago
4 0
It is said to go viral?
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i need help with this my laptop is not let me connet to my wifi and when i try to add out wifi if says "connect Failed" please h
dexar [7]

Answer: try taking it to your media center for help or maybe try connecting to different wifi

Explanation:

7 0
3 years ago
Question 4 1 pts The following code could be rewritten using: if (x > 12): if (x < 34):​
artcher [175]

Answer:

if(x>12 || x<34)

Explanation:

Mashing the two together using the or '||' operator would allow to run both necessary functions for the next code.

7 0
3 years ago
What are the assignable addresses for the 12th subnet with the network address of 220.100.100.0 and the number of needed subnets
bogdanovich [222]

Answer:

C. 220.100.100.45 to 220.100.100.46

Explanation:

The Classless IP subnetting of 220.100.100.0 begins from the fourth octet of the IP address. To get 45 subnet mask, it uses 6 bits from the fourth octet, which approximately give 64 subnets, while the remaining 2 bits are used for host IP addressing.

The useable host IP addresses are gotten from the formula '2^{n}-2', with n=2 bits.

useable host IP addresses = 2^2 - 2 = 2 addresses per subnet.

While the 12th subnet is 12 x 2^2 = 44.

This means that the 12th subnet mask starts with 220.100.100.44 (as the network address) and ends with 220.100.100.47 as broadcast IP address, while '.45' and '.46' are the assignable addresses of the subnet.

5 0
3 years ago
I'm skeptical about (b) , is this accurate or the ranges should be listed differently ?​
Inessa05 [86]

Answer:

My guess would be C2-C11, But I may be wrong

Imagine a graph

Hope this helps....

4 0
3 years ago
Problem 4 (25 points)Consider a byte addressing architecture with 64-bit memory addresses.(a)Which bits of the address would be
zysi [14]

Answer:

Following are the solution to the given points:

Explanation:

The Memory address value = 64 bit

The Size of the word = \frac{64}{8} =8 \  Byte

In point a:

The offset size = 3\ bits ( in 1-word block size)  

The Index size = 9 \  bits (as block number =512)  

Tag size = 64 - 12  = 52\  bits

In point b:

The offset size = 8 \times 8 \ bytes = 2^6 = 6 \ bits.

The Index size = 64 \ bits = 2^6 \ =6 \ bits

Tag size  = 64 - 12  = 52\  bits

In point c:

The Ratio at point a  

\to 3:64

The Ratio at point b

\to 6:64

5 0
3 years ago
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