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3 years ago

Which best describes a difference between transcription and DNA replication

Computers and Technology

2 answers:

ruslelena [56]3 years ago

6 0

DNA replication is just making a copy of someone or somethings DNA

ycow [4]3 years ago

3 0

Answer:

<u>transcription uses uracil. </u>

<u></u>

Explanation:

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Answer:

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3 years ago

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3 years ago

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Which business application uses electronic tags and labels to identify objects wirelessly over short distances?

Varvara68 [4.7K]

Answer:

Option A (Radio Frequency Identification)

Explanation:

Option A (Radio Frequency Identification)

Radio Frequency Identification:

This technology uses radio waves to transfer data between a reader and movable items. The benefit of this technology is we do not require physical contact with items and scanner.

Components of RFID:

tags (computer chip)
Reader
Communication System
RFID software

There are two RFID standards:

Electronic Product Code (EPC) standard
International Standard Organization (ISO)

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3 years ago

A __________ is a repository of data gathered from operational data and other sources that is designed to serve a particular com

VMariaS [17]

Data Mart is a smaller amount of data that is collected from a data warehouse.

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3 years ago

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:

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:

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

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.

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3 years ago