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Masteriza [31]
3 years ago
5

In the wireless telecommunications industry, different technical standards are found in different parts of the world. A technica

l standard known as GSM is common in Europe, and an alternative standard, CDMA, is more common in the United States and parts of Asia. Equipment designed for GSM will not work on a CDMA network and vice versa. What kind of competitive pressure does this create for multinational companies in the industry?
Computers and Technology
1 answer:
Serhud [2]3 years ago
3 0

Answer:

A varying infrastructure.

Explanation:

The competitive pressure it creates for multinational companies in the industry is one of difference in infrastructure because different technical standards are found in different parts of the world.

For instance, A technical standard known as Global Systems for Mobile (GSM) is common in Europe, and an alternative standard, Code Division Multiple Access (CDMA), is more common in the United States and parts of Asia.

Consequently, equipment designed for GSM will not work on a CDMA network and vice versa as a result of varying infrastructure, so companies would be innovative in order to improve on their products.

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The rhythmic note that three beats is called a____half note.
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Answer:

it is called a dotted half note

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the header, in an academic report, typically contains the author’s name and the current page number true or false
kolbaska11 [484]
TRUE

True


In most academic reports, most specifically, an MLA academic report, Student’s last name and current page number is contained in the headers. In MLA, which is most common used formatting guide in academic reports, headers numbers all pages consecutively within the right margin.


6 0
3 years ago
What is the answer to this question?
Ivan

5

8

<u>Explanation:</u>

<u></u>

Since the number is already defined as 5, greater_than_zero holds the value true and less_than_zero holds the value false because 5 is greater than zero and not less than 0. Since there are 4 four if conditions, only two holds true i.e the first one and the last one because the in third condition we are using and logic which needs both condition to be true to make the condition true, whereas in fourth we are using or logic which can make condition true even if one condition is true. Hence the number printed will be number and number +3.

6 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  \\\\

3 0
3 years ago
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