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iren2701 [21]
3 years ago
12

Lin is booting up his computer, and during the boot process, the computer powers down. After several unsuccessful attempts to bo

ot his computer, he decides that he needs to start the troubleshooting process. Lin thinks that the problem could be in the power supply, so he uses a power supply tester. The test shows that the power supply is functioning properly. What should Lin check next in the troubleshooting process?
Computers and Technology
1 answer:
mel-nik [20]3 years ago
3 0

Answer:

The RAM Modules.

Explanation:

If the power supply is working properly, the next thing that could cause an auto-shutdown could be the RAM.

Sometimes static electricity,  a faulty slot, or even a faulty memory module  could be causing the RAM to fail. And as the OS needs to read from the RAM in order to boot, at the moment the processor can't find the RAM information, it shuts down the system.

Changing the RAM modules to a different slot, switching slots, or cleaning the memory module pins with a regular eraser can help solve the problem. If not, then Lin might need to buy a new module, or keep going forward with the diagnostic process.

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The instruction format of a computer is given by one indirect bit, opcode bits, register address bits, immediate operand bits an
oksian1 [2.3K]

Answer:

Following are the answer to this question:

Explanation:

A)

The memory size is 1 Giga Bytes which is equal to 2^{30}

\texttt{Number of address bits    \ \ \ \ \ \ \ \ \ \ \ \ \ \            Number of addresses}\\\\ \ \ \ \ 30  \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \  \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \                                 2^{30}= 1073741824

B) \texttt{I \ \ \ Opcode  \ \ \ Register address \ \ \  Immediate operand \ \ \    Memory address}\\\\\\\textt{1 \ bit  \  \ \ \ \ \ 128  \ \ \ \ \ \ \ \ \ \ \ \ \ \ \    2 \ bits   \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \   24 \ bits  \ \ \ \ \ \ \ \ \ \ \ \ \ \ \  \ \ \ \ \ \ \ \ \           30 \ bits}\\\\\\            = 2^7 \\\\ = 7 \ bits

calculating the register Bits:

= 64-(1+7+24+30)\\\\=64 -62\ \  bits\\\\= 2\ \ bits\\

C)

Immediate value size while merging the additional benefit with the address field:

= 2^{24} + 2^{30}\\\\= 2^{54}\\\\\texttt{Range before combining(-,+) 24 bits   \ \ \ \ \ \       Range after combining( -,+)54bits}\\\\                         \textt{-2^{12} from + (2^{12}-1)   \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \                                    -2^{27} from + (2^{27}- 1)}

= \frac{24}{2} = 12\\\\= \frac{54}{2} = 27

The range is accomplished by dividing the bits by 2 into the two sides of the o and the number is one short to 0.

8 0
3 years ago
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4 years ago
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miskamm [114]
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3 years ago
(display five messages) write a program that displays welcome to java five times.
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A list of multiple choices is given;

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<span>b)      </span>Rent the apartment anyway.

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