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madreJ [45]
3 years ago
15

Pls help will give brainlest​

Computers and Technology
2 answers:
oksian1 [2.3K]3 years ago
7 0

Answer:using a louder voice

Source:trust me bro

Sloan [31]3 years ago
5 0

Answer:

C

Explanation:

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Describe the benefits of having an operational data warehouse (DWH), and the challenges of operating a DWH repository.
kvasek [131]

Answer:

The benefits of having the operational data ware-house are as follows:

  • The single operational warehouse typically support various tactical and the strategical decisions.  
  • It is subject oriented,volatile and integrated data warehousing in the system.
  • The warehousing mainly allow the system to process the large complex data in more efficient manner.
  • The operational warehouse are basically implemented for many business purposes.

The challenges of the operational warehouse repository are as follows:

  • In the operational data warehouse there is large number of user expectations.
  • The cost is the main challenge while designing the operational data warehousing as, it required efficient system in low cost.
  • Choosing the efficient and correct type of the warehouse is the big challenge.

5 0
3 years ago
The roman structure that features heavy use of arches and columns is the _______.
lara31 [8.8K]
The appropriate response is colosseum. It is otherwise called Coliseum and Flavian Amphitheater is an oval amphitheater in the focal point of the city of Rome, Italy. Worked of cement and sand, it is the biggest amphitheater at any point assembled.
8 0
3 years ago
Many interpretations of Hamlet exist, from long to short, from serious to humorous, and each calls attention to different aspect
madam [21]

Answer:

This video presents only the most important points in the plot:

the appearance of Hamlet's father's ghost

Claudius's efforts to determine the cause of Hamlet's strange behavior

Hamlet's soliloquy

Hamlet's rudeness to Ophelia, which leaves her heart-broken

the staging of the play, after which Hamlet is certain of Claudius's guilt

the scene in which Hamlet loses an opportunity to kill Claudius

the scene in which Hamlet rebukes his mother and then accidentally kills Polonius

Claudius's plan to kill Hamlet, which he shares with Laertes

the scene in the graveyard, in which Hamlet holds Yorick's skull

the scuffle between Laertes and Hamlet when Ophelia is buried

the duel between Laertes and Hamlet, followed by Gertrude's death, Hamlet's killing of Claudius, and Hamlet's death

The quick succession of events in this video seems to suggest that the sequence of events in the play is rather unlikely.

The major themes of the play, such as appearance versus reality, moral corruption, the complexity of action, and the mystery of death are not evident in this adaptation. Other important aspects of the play that the video does not address are the question of Hamlet's madness, his internal conflict, and his inaction. The video ignores characterization in Hamlet and the treatment of philosophical ideas throughout the play.

PLATO answer

Explanation:

PLATO said so

6 0
3 years ago
True/false: the cause of failure in a computer program is often due to more than one factor
kozerog [31]
True of course the computer is a big electronic
7 0
3 years ago
Read 2 more answers
(a) How many locations of memory can you address with 12-bit memory address? (b) How many bits are required to address a 2-Mega-
I am Lyosha [343]

Answer:

Follows are the solution to this question:

Explanation:

In point a:

Let,

The address of 1-bit  memory  to add in 2 location:

\to \frac{0}{1}  =2^1  \ (\frac{m}{m}  \ location)

The address of 2-bit  memory to add in 4 location:

\to \frac{\frac{00}{01}}{\frac{10}{11}}  =2^2  \ (\frac{m}{m}  \ location)

similarly,

Complete 'n'-bit memory address' location number is = 2^n.Here, 12-bit memory address, i.e. n = 12, hence the numeral. of the addressable locations of the memory:

= 2^n \\\\ = 2^{12} \\\\ = 4096

In point b:

\to Let \  Mega= 10^6

              =10^3\times 10^3\\\\= 2^{10} \times 2^{10}

So,

\to 2 \ Mega =2 \times 2^{20}

                 = 2^1 \times 2^{20}\\\\= 2^{21}

The memory position for '2^n' could be 'n' m bits'  

It can use 2^{21} bits to address the memory location of 21.  

That is to say, the 2-mega-location memory needs '21' bits.  

Memory Length = 21 bit Address

In point c:

i^{th} element array addresses are given by:

\to address [i] = B+w \times (i-LB)

_{where}, \\\\B = \text {Base  address}\\w= \text{size of the element}\\L B = \text{lower array bound}

\to B=\$ 52\\\to w= 4 byte\\ \to L B= 0\\\to address  = 10

\to address  [10] = \$ 52 + 4 \times (10-0)\\

                       =   \$ 52   + 40 \ bytes\\

1 term is 4 bytes in 'MIPS,' that is:

= \$ 52  + 10 \ words\\\\ = \$ 512

In point d:

\to  base \ address = \$ t 5

When MIPS is 1 word which equals to 32 bit :

In Unicode, its value is = 2 byte

In ASCII code its value is = 1 byte

both sizes are  < 4 byte

Calculating address:

\to address  [5] = \$ t5 + 4 \times (5-0)\\

                     = \$ t5 + 4 \times 5\\ \\ = \$ t5 + 20 \\\\= \$ t5 + 20  \ bytes  \\\\= \$ t5 + 5 \ words  \\\\= \$ t 10  \ words  \\\\

3 0
3 years ago
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