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IRINA_888 [86]
3 years ago
14

You are the manager and an employee shows up for his shift dressed inappropriately for work. What should you do?

Computers and Technology
2 answers:
yarga [219]3 years ago
7 0

Most managers see inappropriate dressing every day. Some are left angered while some are confused as to what is the right way to approach such situations. What follows next needs to be professional oriented. Being the manager, I would start by finding out what the appropriate dressing policies are. Some companies might have dress code policies while others might not. Assuming that this company has, I would read these policies carefully. The next thing that is required from me is to have a small conversation with the employee discretely. As professional as I can be, I will mention how I feel his or her dress code is inappropriate and try to summon the employee by asking him or her whether they are fully aware of the appropriate dress code. I will try to agree with the employee on what is appropriate and what is not. If we agree, requesting the employee to home and change clothes is unnecessary, but if the need to do that arises, I would consider that to be an option.

olga2289 [7]3 years ago
6 0

this is late, but for future people, the answer is B

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Explanation:

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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

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A computer hard disk starts from rest, then speeds up with an angular acceleration of 190 rad/s2 until it reaches its final angu
Nataly_w [17]
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 θ1 = 0 + (1/2) * (190) * (3.97) ^ 2
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