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3 years ago

Write an algorithm to show whether a given number is even or odd.

1 answer:

5 0

Answer:
Step 1: Start
Step 2: Take any number and store it in n
Step 3: if n=multiple of 2 print "even" else print "odd"
Step 4: Stop
Explanation:
If the number is dividable or a multiple of 2, the number is even. If it's not the number is odd

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......................... and ....................... are two ways to define the text in Inkscape.

crimeas [40]

Answer:

There are two ways to add text to an Inkscape drawing. The first is as regular text. In this case, as text is typed, the text box grows to accommodate the text. Line breaks must be manually added.

Explanation:

4 0

3 years ago

(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

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MA_775_DIABLO [31]

Computers have changed the way people relate to one another and their living environment, as well as how humans organize their work, their communities, and their time. Society, in turn, has influenced the development of computers through the needs people have for processing information.

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3 years ago

What happens if the cursor is in the last cell in a table and you press Tab?

Mashutka [201]

I think it would move to the beginning of the next row

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4 years ago

What technology will examine the current state of a network device before allowing it can to connect to the network and force an

miv72 [106K]

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3 years ago