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3 years ago

The data table provides the results of several what-if analyses, but results are often clearer if you include a(n

1 answer:

4 0

CVP Chart.
Source: "New Perspectives on Microsoft Excel 2010: Comprehensive" by June Parsons, Dan Oja, Roy Ageloff, Patrick Carey.

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Write a SELECT statement without a FROM clause that uses the NOW function to create a row with these columns: today_unformatted

OlgaM077 [116]

Answer:

SELECT

NOW() AS 'today_unformatted',

DATE_FORMAT(NOW(), '%d-%b-%Y') AS 'today_formatted';

Explanation:

%d represents date.

%b represents month.

%Y represents year.

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3 years ago

_______ is the type of vision useful in sensing motion and objects outside normal vision.

Likurg_2 [28]

The answer is peripheral vision, also known as side vision.

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3 years ago

IF ANYONE USES THE COMPUTER PROGRAM SCRATCH I NEED YOUR HELP!! 15 POINTS!!

Irina-Kira [14]

The mouse down block C, 2) is c, and

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4 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:

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

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

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4 years ago

When an individual user works in _____, the workstation performs all the functions of a server by storing, accessing, and proces

anyanavicka [17]

When an individual user works in stand-alone mode, the workstation performs all the functions of a server by storing, accessing, and processing data as well as providing a user interface.

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3 years ago