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ANEK [815]
3 years ago
14

Most printers are plug and play compatible and must be manually configured when they are plugged into the system.

Computers and Technology
2 answers:
nikdorinn [45]3 years ago
8 0

Answer:

False

Explanation:

Plug and play devices are computer peripheral devices that can be used immediately, with little or no necessary configuration when plugged or connected to the computer system.

Printers are mostly electronic devices used as an output of a hard copy of a computer system application document. They are not part of the computer system (peripheral devices). They come with installation disk which a computer must install, in order to be able to use the printer.

Yes they are peripheral devices, but they are not plug and play devices since its software must be installed on the computer, to use it

Morgarella [4.7K]3 years ago
3 0

Answer:

The statement is: False.

Explanation:

Plug and Play or PnP devices are those which by just connecting them into a server are immediately recognized by the Operating System (OS). Though, some other devices like, printers, and additional monitors need a software (driver) to be installed so they can be fully used.

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What is the correct sequence in which a computer operates​
GrogVix [38]

Answer:

Booting is a startup sequence that starts the operating system of a computer when it is turned on. A boot sequence is the initial set of operations that the computer performs when it is switched on. Every computer has a boot sequence.

8 0
3 years ago
A person you respect who offers you advice and assistance is known as a(n) _____.
Archy [21]
This would be a mentor-D. They role is to mentor or guide you with advice or assistance. A role model is a similar answer in that they respect you, but they might not offer advice or assistance which makes it incorrect.
7 0
3 years ago
Translate the following MIPS code to C. Assume that the variables f, g, h, i, and j are assigned to registers $s0, $s1, $s2, $s3
Romashka [77]

Answer:

f = 2 * (&A[0])

See explaination for the details.

Explanation:

The registers $s0, $s1, $s2, $s3, and $s4 have values of the variables f, g, h, i, and j respectively. The register $s6 stores the base address of the array A and the register $s7 stores the base address of the array B. The given MIPS code can be converted into the C code as follows:

The first instruction addi $t0, $s6, 4 adding 4 to the base address of the array A and stores it into the register $t0.

Explanation:

If 4 is added to the base address of the array A, then it becomes the address of the second element of the array A i.e., &A[1] and address of A[1] is stored into the register $t0.

C statement:

$t0 = $s6 + 4

$t0 = &A[1]

The second instruction add $t1, $s6, $0 adding the value of the register $0 i.e., 32 0’s to the base address of the array A and stores the result into the register $t1.

Explanation:

Adding 32 0’s into the base address of the array A does not change the base address. The base address of the array i.e., &A[0] is stored into the register $t1.

C statement:

$t1 = $s6 + $0

$t1 = $s6

$t1 = &A[0]

The third instruction sw $t1, 0($t0) stores the value of the register $t1 into the memory address (0 + $t0).

Explanation:

The register $t0 has the address of the second element of the array A (A[1]) and adding 0 to this address will make it to point to the second element of the array i.e., A[1].

C statement:

($t0 + 0) = A[1]

A[1] = $t1

A[1] = &A[0]

The fourth instruction lw $t0, 0($t0) load the value at the address ($t0 + 0) into the register $t0.

Explanation:

The memory address ($t0 + 0) has the value stored at the address of the second element of the array i.e., A[1] and it is loaded into the register $t0.

C statement:

$t0 = ($t0 + 0)

$t0 = A[1]

$t0 = &A[0]

The fifth instruction add $s0, $t1, $t0 adds the value of the registers $t1 and $t0 and stores the result into the register $s0.

Explanation:

The register $s0 has the value of the variable f. The addition of the values stored in the regsters $t0 and $t1 will be assigned to the variable f.

C statement:

$s0 = $t1 + $t0

$s0 = &A[0] + &A[0]

f = 2 * (&A[0])

The final C code corresponding to the MIPS code will be f = 2 * (&A[0]) or f = 2 * A where A is the base address of the array.

3 0
3 years ago
Anyone help me with number 41.
Katarina [22]
It is te = t I yhink
8 0
3 years ago
What is the name of the file in which this
anyanavicka [17]

Answer:

B

Explanation:

i’m an information technology  student and as I remember we used this  name

3 0
3 years ago
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