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Ray Of Light [21]
3 years ago
11

Which of the following can be stored in a computer file?

Computers and Technology
1 answer:
Fed [463]3 years ago
8 0

Answer:

No image uploaded so there can not be an answer provided by the community. Retry uploading the image?

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Learning Management Systems (LMS) allow students to interact with each other, but NOT their teachers.
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what do you mean friend not for teacher

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Assume there is a class AirConditioner that supports the following behaviors: turning the air conditioner on and off, and checki
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yes?

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3 years ago
The publisher tab in the application control allows you to manage the various certificates that are used to do what to binaries?
artcher [175]

The publisher tab in the application control allows you to manage the various certificates that are used to sign to binaries. Publisher allow executables of a particular vendor, signed by a security certificate issued to the vendor by a Certificate Authority. Also, all applications and binary files either added to or modified on an endpoint that are signed by the certificate are automatically added to the whitelist.

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3 years ago
A computer is not connecting to certain network devices properly. the ip address information is as follows: ip address: 192.168.
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192.168 suggests class C networks which have a 24 bit netmask (255.255.255.0) but you haven't provided enough info, like the netmask of a working machine, to be definite.
3 0
3 years ago
Suppose you have a 16-bit machine with a page size of 16B. Assume that any unsigned 16-bit integer can be a memory address. Also
irinina [24]

Answer:

2^11

Explanation:

Physical Memory Size = 32 KB = 32 x 2^10 B

Virtual Address space = 216 B

Page size is always equal to frame size.

Page size = 16 B. Therefore, Frame size = 16 B

If there is a restriction, the number of bits is calculated like this:  

number of page entries = 2^[log2(physical memory size) - log2(n bit machine)]

where

physical memory size = 32KB  which is the restriction

n bit machine = frame size = 16

Hence, we have page entries = 2^[log2(32*2^10) - log2(16)] = 2ˆ[15 - 4 ] = 2ˆ11

7 0
3 years ago
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