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ankoles [38]
3 years ago
10

C#

Computers and Technology
1 answer:
Shkiper50 [21]3 years ago
5 0

Answer:

Hhhhhhjjjjjjkiiiiioooooooooioi

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Crashing almost always accelerates the schedule while increasing project risk, whereas fast tracking almost always accelerates t
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Explanation: Because fast-tracking doesn’t involve additional resources, it’s usually best to start with this technique while Crashing requires careful planning to prevent “panic mode” among the project crews and to avoid acceleration claims.

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Plz..... Add the following Binary numbers 1. 111000 + 1100 + 111111​
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11000011

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Imagine a typical website that works as a storefront for a business, allowing customers to browse goods online, place orders, re
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Following the information given in the question, the testing process for the website will include testing the links that are on the site.

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8.3 code practice edhesive PLEASE HELP AND HURRY
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Answer:

numbers = '14 36 31 -2 11 -6'

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8 0
2 years ago
1. How many bits would you need to address a 2M × 32 memory if:
Dominik [7]

Answer:

  1. a) 23       b) 21
  2. a) 43        b) 42
  3. a) 0          b) 0

Explanation:

<u>1) How many bits is needed to address a 2M * 32 memory </u>

2M = 2^1*2^20, while item =32 bit long word

hence ; L = 2^21 ; w = 32

a) when the memory is byte addressable

w = 8;  L = ( 2M * 32 ) / 8 =  2M * 4

hence number of bits =  log2(2M * 4)= log2 ( 2 * 2^20 * 2^2 ) = 23 bits

b) when the memory is word addressable

W = 32 ; L = ( 2M * 32 )/ 32 = 2M

hence the number of bits = log2 ( 2M ) = Log2 (2 * 2^20 ) = 21 bits

<u>2) How many bits are required to address a 4M × 16 main memory</u>

4M = 4^1*4^20 while item = 16 bit long word

hence L ( length ) = 4^21 ; w = 16

a) when the memory is byte addressable

w = 8 ; L = ( 4M * 16 ) / 8 = 4M * 2

hence number of bits = log 2 ( 4M * 2 ) = log 2 ( 4^1*4^20*2^1 ) ≈ 43 bits

b) when the memory is word addressable

w = 16 ; L = ( 4M * 16 ) / 16 = 4M

hence number of bits = log 2 ( 4M ) = log2 ( 4^1*4^20 ) ≈ 42 bits

<u>3) How many bits are required to address a 1M * 8 main memory </u>

1M = 1^1 * 1^20 ,  item = 8

L = 1^21 ; w = 8

a) when the memory is byte addressable

w = 8 ; L = ( 1 M * 8 ) / 8 = 1M

hence number of bits = log 2 ( 1M ) = log2 ( 1^1 * 1^20 ) = 0 bit

b) when memory is word addressable

w = 8 ; L = ( 1 M * 8 ) / 8 = 1M

number of bits = 0

5 0
2 years ago
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