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const2013 [10]
3 years ago
8

Conner has located a research source that is sponsored by a pharmaceutical company. Based on the sponsorship, Conner must be car

eful that the information is not
Computers and Technology
1 answer:
gogolik [260]3 years ago
6 0

Answer:

The answer to this question is given below in the explanation section.

Explanation:

The correct answer is:

Conner has located a research source that is sponsored by a pharmaceutical company. Based on the sponsorship, Conner must be careful that the information is not <u>Biased.</u>

<u></u>

When you evaluate and use information from multiple sources, you need to evaluate the information that it must be error-free and not biased and it should be accurate information as you needed.

As in this example, Conner needs to research that the information should not be biased because the information or research source is sponsored by a pharmaceutical company and there is chances of biasedness. So, Conner must be careful that information is not biased.

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Answer:

all of the above

Explanation:

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(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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Create a file named network.py. The file should be created in the same directory as health_checks.py, i.e., scripts. If you are
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What is the purpose of the boot loader during the bootup process
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hope this helps 
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Corinne should scroll down to the bottom of the page to where she can see the link
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