Computers and Technology, 25.11.2020 14:10 estherboocx

At a minimum, how many bits are needed in the MAR with each of the following memory sizes? a. 1 million bytes
b. 10 million bytes
c. 100 million bytes
d. 1 billion bytes

Answers: 1

Show answers

Answers

Answer from: babra

Using the bit-byte relationship, the minimum bit memory allocation for the memory sizes stated are 20, 24, 27 and 30 bits respectively.

To obtain the minimum number of bits required for each of the given memory sizes :

We use the relation :

$2^{n}$

Where, n = number of bits

For 1 million byte :

$2^{n} = 1048576$

$2^{n} = 2^{20}$

n = 20

2.) For 10 million bytes :

$2^{n} ≥ 10,000,000$

$2^{24} = 16777216$

All integers before 24 equates to a value less than 10,000,000

n = 24

3.) For 100 million bytes :

$2^{n} ≥ 100,000,000$

$2^{27} = 134217728$

n = 27

All integers before 27 equates to a value less than 100,000,000

4. For 1,000,000,000 bytes :

$2^{n} = 1073741824$

$2^{30} = 2^{30}$

n = 30

All integer values before 30 equated to a value less than 1,000,000,000

Learn more :link

Answer from: bigJ4864

a. 20 bits

b. 24 bits

c. 27 bits

d. 30 bits

Explanation:

Using base 2 numbers where a location in RAM is represented using an address.

Hence, given that with 2 bits one can store 4 things maximum, with 3 bits, one can store between 5 and 8 things.

A. Therefore to represent 1 million bytes -1 addresses, we need 20 bits. This depicts 1,048,576 to be exact. This is a megabyte or 1 MB.

B For 10 million bytes we have 1 million different addresses using 20 bits.

This equates to 10 million is 10 x bigger than 2^20.

With each addition of one bit to the address, the addresses to represents doubled e double the number of addresses that we can represent (and reach).

For instance : 1 million -> 2 million -> 4 million -> 8 million -> 16 million = 2^20 -> 2^21 -> 2^22 -> 2^24

Thus, 24 bits equates to 10 million addresses. 2^23 < 10 million < 2^24 where 2^24 = 16,777,216.

C. In 100 million bytes, 27 bits are required.

Thereby 2^26 = 67,108,864 bytes, not enough for 100 million. 2^27 = 134,217,728 and this is enough.

D. 1 billion bytes is equal to 1000 times 1 million or 1000 * 2^20.

Therefore in base 2 notation, it is (2^20 * 2^10) = 2^30 = 1,073,741,824.

Hence 30 bits are required to represent 1 billion bytes. This equates to 1GB.

Answer from: Quest

answerequal fluid intake and fluid loss;

Answer from: Quest

Answera fixed regular payment;

Another question on Computers and Technology

Computers and Technology, 21.06.2019 20:30

What important technology has done the most to allow a businesses a chance to compete with larger international companies

Answers: 1

Answer

Computers and Technology, 21.06.2019 23:10

Write a method that accepts a string object as an argument and returns the number of words it contains. for instance, if the argument is "four score and seven years ago", the method should return the number 6. demonstrate the method in a program that asks the user to input a string and then passes that string into the method, printing out whatever the method returns.

Answers: 3

Answer

Computers and Technology, 22.06.2019 11:30

Awell-diversified portfolio needs about 20-25 stocks from different categories is this true or false?

Answers: 2

Answer

Computers and Technology, 22.06.2019 17:20

[a] create a class called “cycle” which has two instance integer variables as properties, “numberofwheels” and “weight.” create a constructor with two parameters, using the same variable names in the parameter list. assign each variable to numberofwheels” and “weight” respectively. write a separate application to test the class and display its properties. note: do not change the names of the instance variables or the variables listed in the constructor’s parameter list. [b] edit your class cycle by adding a default constructor which will assign the default values of 100 to represent the numberofwheels, and 1000 to represent the weight, by invoking a call to the other constructor. modify your application created in [a] to test the class.

Answers: 3

Answer

You know the right answer?

At a minimum, how many bits are needed in the MAR with each of the following memory sizes? a. 1 mil...

Questions