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nignag [31]
3 years ago
8

Most network behavior analysis system sensors can be deployed in __________ mode only, using the same connection methods as netw

ork-based idpss.
Computers and Technology
1 answer:
soldier1979 [14.2K]3 years ago
7 0
Most network behavior analysis system sensors can be deployed in passive mode only, using the same connection methods as network based idpss. Idps works as a network or host based systems. Most Nba sensors can be deployed in passive mode only too. The Nba examines the network traffic.
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Explanation:

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3 years ago
When replacing a system board in a laptop, which feature is a must?
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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
3 years ago
Given a positive real number, print its fractional part.
givi [52]
<span>The modf() function will do this for you:

double x, y, d;
x = -14.876;
y = modf(x, &d);
printf("Fractional part = %lf\n", y);

</span>
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Which formulas would work to combine cells with first, middle, and last names from columns A through C and row 2 into a new cell
AURORKA [14]

Split? not really sure but i understand a bit of this..

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