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serg [7]
3 years ago
14

In an array based implementationof a queue a possible solution to dealing with the full condition is to

Computers and Technology
1 answer:
kati45 [8]3 years ago
6 0

Answer:

D. Maintain a count of queue items.

Explanation:

For checking the full condition of a queue you have to  check whether the number of elements in the array are equal to the size of the array.If they are equal then we can that the queue is full.

Checking frontindex equal to backindex is for checking if the queue is empty.

In option C we have to check for arrayFullException but it will not be a good approach.

So we conclude by saying that the answer is option D.

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The hexadecimal number system uses alphabets A to F to represent values_ to _
mash [69]

Answer:

they represent values 10 to 15

7 0
3 years ago
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By generating and delivering timely and relevant information supported by networks, _____ creates new opportunities for conducti
castortr0y [4]

Answer:

The correct option to the following question is option (B). e-business.

Explanation:

E-business is stands for the Electronic business.

E-business is the business which is conduct by the uses of the internet, Web, extranet or intranet, etc.

E-business is also known as the online business where online transactions take place.

E-business facilitates our customers that the selling and the buying of the goods between the consumers and the merchants.

7 0
3 years ago
Please help on both of these questions will give brainliest<br> !!
Alja [10]

I think the first one is All of the above

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3 years ago
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Jean has created a database to track all of this month’s family meals. To find an entry for lasagna, what would should Jean do?
Nutka1998 [239]
A or d


A edit data

And
.
,
D sort of data
6 0
3 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
3 years ago
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