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kifflom [539]
3 years ago
14

You enter the show ipv6 route command on an ospf device and the device displays a route. which conclusion can you draw about the

environment
Computers and Technology
1 answer:
salantis [7]3 years ago
4 0
<span>OSPFv3 is in use.......</span>
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Why do some people put salt on sidewalks when it snows?
drek231 [11]
It help to melt the snow quicker. they salt eats away the snow.
6 0
3 years ago
What are the pros and cons of MP3 audio archives?
UkoKoshka [18]
Was this in reference to literal audio archives? If so, I don't see any cons beside possible copyright infringement.

If you're talking about the codecs themselves, then I can do that.

<span>Pros:

</span>- Widespread acceptance. Supported in nearly all hardware devices, and continually adopted by newer ones.

- Faster decoding. Much more so than FLAC, Vorbis, etc.

- Relaxed licensing schedule.

<span>Cons:
</span><span>
</span>- Lower quality and efficiency than most modern codecs. (To be fair, never really noticed this one).

- Sometimes the maximum bitrate isn't enough.

- Pretty much void/unusable for high definition audio (higher than <span>48kHz).</span>

 
7 0
3 years ago
Write a program to test the difference between %d and %i conversion
Gelneren [198K]
Void test(char *s)
{
  int i, d;
  sscanf(s, "%i", &i);
  printf("%s converts to %i using %%i\n", s, i);
  sscanf(s, "%d", &d);
  printf("%s converts to %d using %%d\n", s, d);
}

int main()
{
  test("123");
  test("0x123");
  return 0;
}

outputs:
123 converts to 123 using %i
123 converts to 123 using %d
0x123 converts to 291 using %i
0x123 converts to 0 using %d

As you can see, %i is capable of parsing hexadecimal, whereas %d is not. For printf they're the same.
7 0
3 years ago
A computer has a cache and main memory. 20 ns are required to reference a word that is in the cache. If the word is in main memo
Alecsey [184]

Answer:

T_{average} =12026ns

Explanation:

Given Information:

T_{cache}=20 ns

T_{main}=60ns

T_{disk} =12ms=12,000,000ns

hit_{cache} =0.9

miss_{cache} =1-0.9=0.10

hit_{main} =0.99

miss_{main} =1-0.99=0.01

There are 3 possible cases

1. Location of reference word is in cache

T_{cache}*hit_{cache} =20*0.9=18ns

2. Location of reference word is not in cache but in main memory

(T_{cache}+T_{main})*(miss_{cache}*hit_{main}) =(20+60)*(0.10*0.99)=7.92ns

3. Location of reference word is neither in cache nor in main memory

(T_{cache}+T_{main}+T_{disk})*(miss_{cache}*miss_{main})

(20+60+12,000,000)*(0.10*0.01)=12000.08 ns

The average time required to access a referenced word on this system is simply the sum of above 3 cases

T_{average} =18+7.92+12000.08=12026ns

5 0
3 years ago
What is word processing in excel
Flura [38]

Answer:

Microsoft Excel and Microsoft word are the part of the MicrosoftOffice Package of productvity software.

5 0
2 years ago
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