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Ymorist [56]
3 years ago
7

A n n sss eeeee rrrr

Computers and Technology
1 answer:
Nadusha1986 [10]3 years ago
4 0

Usability emphasizes utility over aesthetics because in order for something to be useful it has to have a designated purpose. While aesthetics are nice they are not usable so the answer is B

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3. Which swim_backwards method is called when sammy.swim_backwards() is executed? Why?
Georgia [21]
Classes called child classes or subclasses inherit methods and variables
4 0
2 years ago
Consider sending a 10000-byte datagram into a link that has an MTU of 4468 bytes. Suppose the original datagram is stamped with
AfilCa [17]

Answer:

Number of fragments is 3

Explanation:

The maximum size of data field in each fragment = 4468 - 20(IP Header)

= 4448 bytes

Hence, the number of required fragment = (10000 - 20)/4448

= 3

Fragment 1

Id = 218

offset = 0

total length = 4468 bytes

flag = 1

Fragment 2

Id = 218

offset = 556

total length = 4468 bytes

flag = 1

Fragment 3

Id = 218

offset = 1112

total length = 1144 bytes

flag = 0

5 0
2 years ago
You listened to a song on your computer. did you use hardware or software? explain.
scZoUnD [109]
Software, unless you planned on permanently downloading the Music Album or Song to your computer's Hard Drive. i.e, you would use whatever Music or Media player you have installed on your computer, and that simply counts as Software. 
4 0
3 years ago
What are the best 3 xbox 360 games ?​
Andru [333]
Halo
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Call of duty
5 0
3 years ago
Read 2 more answers
) Consider a router that interconnects four subnets: Subnet 1, Subnet 2, Subnet 3 and Subnet 4. Suppose all of the interfaces in
erik [133]

Answer:

Check the explanation

Explanation:

223.1.17/24 indicates that out of 32-bits of IP address 24 bits have been assigned as subnet part and 8 bits for host id.

The binary representation of 223.1.17 is 11011111 00000001 00010001 00000000

Given that, subnet 1 has 63 interfaces. To represent 63 interfaces, we need 6 bits (64 = 26)

So its addresses can be from 223.1.17.0/26 to 223.1.17.62/26

Subnet 2 has 95 interfaces. 95 interfaces can be accommodated using 7 bits up to 127 host addresses can represented using 7 bits (127 = 27)

and hence, the addresses may be from 223.1.17.63/25 to 223.1.17.157/25

Subnet 3 has 16 interfaces. 4 bits are needed for 16 interfaces (16 = 24)

So the network addresses may range from 223.1.17.158/28 to 223.1.17.173/28

4 0
3 years ago
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