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Artemon [7]
2 years ago
6

If you were a hackathon team manager, how could you best address the conflict created by having more volunteers than open roles

Computers and Technology
1 answer:
Nuetrik [128]2 years ago
5 0

There are various ways to resolve conflict. Addressing conflict by the  team manage is by:

  • Control conflict by expanding the resource base
  • Eliminate conflict by assigning volunteers to another project team.

There are a lot of ways an individual can handle conflict. One of which is to expand the resource base.

Example: Supporting  team manager receives 4 budget requests for $150,000 each. If he has only $200,000 to distribute, there will be conflict at this stage because each group will believe its proposal is worth funding and will not be happy if not fully funded. So the best thing to do is to expand the resources allocated.

The right way to fix this conflict is to tell your volunteers in advance that some of them will not be doing different tasks etc.

See full question below

  • Capital One invites employees to work on special projects during hackathons. Employees can explore new technologies, rapidly push development forward, or just expand their network to include more colleagues interested in innovation. There's always the possibility that more employees want to participate in the hackathon than there are roles available. If you were a hackathon team manager, how could you best address the conflict created by having more volunteers than open roles?
  • Control conflict by expanding the resource base.
  • Eliminate conflict by assigning volunteers to another project team.
  • Eliminate conflict by avoiding the volunteers.
  • Stimulate conflict by making volunteers compete for the available roles.

Learn more from

brainly.com/question/18103914

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Levi wants to run 5 commands sequentially, but does not want to create a shell script. He knows that each command is going to ta
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3 0
3 years ago
a password to a certain database consists of digits that cannot be repeated. if the password is known to consist of at least 8 d
BaLLatris [955]

Answer:

\frac{10!}{2}mins

Explanation:

12 seconds to try one combination will be equivalent to  \frac{1}{12}\times 60 = \frac{1}{5} \ mins

Password contain at least 8 digit i.e. password can contain 8, 9, 10 digit.

Password cannot contain more than 10 digit because it will give room for repetition which it is clearly stated that digit cannot be repeated.

Possible digit that can be used: 9,8,7,6,5,4,3,2,1,0.

Total number of passwords combination possible for each position in 8 digit.

1st position = 10, 2nd position = 9, 3rd position = 8, 4th position = 7, 5th position = 6, 6th position = 5, 7th position = 4, 8th position = 3. Total number of passwords combination possible in 8 digit is equivalent to \frac{10!}{2}.

Total number of passwords combination possible for each position in 9 digit.  

1st position = 10, 2nd position = 9, 3rd position = 8, 4th position = 7, 5th position = 6, 6th position = 5, 7th position = 4, 8th position = 3, 9th position = 2. Total number of passwords combination possible in 9 digit is equivalent to \frac{10!}{1}.

Total number of passwords combination possible for each position in 10 digit.

1st position = 10, 2nd position = 9, 3rd position = 8, 4th position = 7, 5th position = 6, 6th position = 5, 7th position = 4, 8th position = 3, 9th position = 2, 10th position = 1.  Total number of passwords combination possible in 10 digit is equivalent to 10!.

Adding them up and multiplying by  \frac{1}{5} \ mins  to get the total number of time needed to guarantee access to database =  [\frac{10!}{2}\times\ \frac{10!}{1} \times 10!] \frac{1}{5}\ mins = \frac{10!}{2}

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
3 years ago
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