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sleet_krkn [62]
3 years ago
13

Question 2 of 10

Computers and Technology
1 answer:
Minchanka [31]3 years ago
7 0

True because it depends

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In the world of computing,accessibility MOST often refers to what
Nutka1998 [239]

<span>Accessibility refers to access to some form of computer technology. This means that numbers are easily organized and data is easily calculated making it easier for people’s lives. It is a means of an efficient and more productive existence in the work environment.</span>

5 0
3 years ago
With landscape photography, which depth of field do you normally want? A) Small B)Small to Medium C)Large D)With landscapes it d
RideAnS [48]
I am not 100% on this but i think you would want a Large depth of field. 
3 0
4 years ago
Read 2 more answers
In python, what is the difference between a dictionary and a set? How are they similar?
yuradex [85]

Answer:

"A set is an unordered collection. A dictionary is an unordered collection of data that stores data in key-value pairs."

Explanation:

Set =>

Collection of non-repetitive elements.

Unordered

Unindexed

No way to change items.

Dictionary =>

Collection of key-value pairs.

Unordered

Indexed

Mutable

4 0
2 years ago
In the following cell, we've loaded the text of Pride and Prejudice by Jane Austen, split it into individual words, and stored t
Sever21 [200]

Answer:

Explanation:

Since the array is not provided, I created a Python function that takes in the array and loops through it counting all of the words that are longer than 5. Then it returns the variable longer_than_five. To test this function I created an array of words based on the synapse of Pride and Prejudice. The output can be seen in the attached picture below.

def countWords(p_and_p_words):

   longer_than_five = 0

   for word in p_and_p_words:

       if len(word) > 5:

           longer_than_five += 1

   return longer_than_five

7 0
3 years ago
Consider sending a 2400-byte datagram into a link that has an mtu of 700 bytes. suppose the original datagram is stamped with th
Feliz [49]

Explanation:

Let, DG is the datagram so, DG= 2400.

Let, FV is the Value of Fragment and F is the Flag and FO is the Fragmentation Offset.

Let, M is the MTU so, M=700.

Let, IP is the IP header so, IP= 20.

Let, id is the identification number so, id=422

Required numbers of the fragment = [\frac{DG-IP}{M-IP} ]

Insert values in the formula = [\frac{2400-20}{700-20} ]

Then,        = [\frac{2380}{680} ] = [3.5]

The generated numbers of the fragment is 4

  • If FV = 1 then, bytes in data field of DG= 720-20 = 680 and id=422 and FO=0 and F=1.
  • If FV = 2 then, bytes in data field of DG= 720-20 = 680 and id=422 and FO=85(85*8=680 bytes) and F=1.
  • If FV = 3 then, bytes in data field of DG= 720-20 = 680 and id=422 and FO=170(170*8=1360 bytes) and F=1.
  • If FV = 4 then, bytes in data field of DG= 2380-3(680) = 340 and id=422 and FO=255(255*8=2040 bytes) and F=0.

3 0
4 years ago
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