<span>Store programming components in the shared folder. you want to secure the data in the folder as follows: members of the research resources group should have full control over the folder. all other users should be able to view the contents of the folder. permissions should apply to both network access through the share and local access. your task in this lab is to implement a combined share and NTFS permission strategy as follows:
1.share the h:\components folder using the default share name.
2.configure share permissions for the share to give the everyone group allow full control share permissions.
3.configure NTFS permissions for the h:\components folder to give the research resources group allow full control permissions.
4.keep the remaining NTFS permissions. this allows administrators full control and allows all other users to read the folder contents.</span>
Answer:
Authorization levels (privileges) need to be set for the users in the company.
Explanation:
Users need to be classified according to their departments and their position. Each group should see only <em>related</em> data, so their authorizations need to be done accordingly.
Also managers and staff should have different levels of reach to the data. Thus their privileges should set accordingly.
There's definitely something wrong with your for loop. Try this:
for x in range(n):
You might want to check if all your print statements are correct. You might be missing just a period and it will be counted wrong.
As we know that 10-bAse data travels at the rate of 10 mbps,
therefore at 1 sec we have 10240 bytes.
We have to calculate for 64 bytes travelling in one second.
Multiply both sides by 64.
64 sec = 10240 x 64 bytes
64 bytes = 64 / 10240 sec
Now we have to calculate for light of speed
1 sec = 186000 miles
Substitute this value to the formula above
64 bytes = 64 / 10240 x 1 sec = 64 / 10240 x 186000 miles =
64 / 10240 x 186000 x 5280 feet = 6138x10^3 feet
Therefore, 64 bytes is equal to 6138 x 10^3 feet long
message
Answer:
For values when n<32 use fb(n) else use fa(n).
See explaination for details
Explanation:
Earlier when n=1 fa(n)=2000 while fb(n)=2. for n=2 fa(n)=2000*22 =2000*4=8000 while fb(n)=2*24=2*16=32. It is observed that fa(n) requires more time than fb(n) for small values.
Now, we will see when fb(n) crosses fa(n). This can happen only when fb(n) values equals or greater than fa(n)
therefore,
2000n2<=2n4
Solving equation we get n2>=1000 which can happen when n>=32.
So for values when n<32 use fb(n) else use fa(n)
