To fetch data from secondary memory, certain registers are used to handle the addresses and manage the data flow between secondary memory and the CPU.
The register used to fetch data from secondary memory is the $\boxed{\text{Memory Address Register (MAR)}}$.
Consider the following statements about the range of numbers in a $9$-bit $1$'s complement and $2$'s complement system.
I. In $9$-bit $1$'s complement, the range is $-255$ to $+255$, and there exist two representations of zero.
II. In $9$-bit $2$'s complement, the range is $-256$ to $+255$, and both $1$'s complement and $2$'s complement can represent exactly $512$ unique values.
III. The maximum positive number representable is $+255$ in both $1$'s complement and $2$'s complement $9$-bit systems.
Identify the CORRECT option.
For $9$-bit $1$'s complement, the range is
$-(2^{8}-1)$ to $+(2^{8}-1)$
$=-255$ to $+255$
Also, in $1$'s complement, there are two representations of zero: positive zero and negative zero.
So, statement I is correct.
For $9$-bit $2$'s complement, the range is
$-2^8$ to $2^8-1$
$=-256$ to $+255$
But $1$'s complement does not represent exactly $512$ unique values because zero has two representations.
So, statement II is incorrect.
The maximum positive number in both systems is
$+255$
So, statement III is correct.
Therefore, statements I and III only are correct.
Binary multiplication is done similarly to decimal multiplication, where each bit is multiplied individually. Let's break down \( 00_2 \times 11_2 \) step by step.
We multiply each digit in the first binary number by each digit in the second binary number. The multiplication follows the same rules as decimal multiplication but with only 0's and 1's.
Let's multiply the two binary numbers:
Since all the results are 0, the final multiplication result is also 0.
The result of \( 00_2 \times 11_2 \) is: 0
First convert both binary numbers into decimal.
$1100_2=12$
$1011_2=11$
Now multiply:
$12\times 11=132$
Convert $132$ into binary:
$132=128+4$
$128=2^7$ and $4=2^2$
So,
$132=10000100_2$
Therefore,
$1100_2\times 1011_2=10000100_2$
Given expression is
$(x+y'+z')(x+y'+z)(x+y+z')$
Using the identity
$(x+A)(x+B)=x+AB$
First take
$(x+y'+z')(x+y'+z)$
Here,
$A=y'+z'$ and $B=y'+z$
So,
$(x+y'+z')(x+y'+z)=x+(y'+z')(y'+z)$
Now,
$(y'+z')(y'+z)=y'+z'z$
Since
$z'z=0$
So,
$(y'+z')(y'+z)=y'$
Now the expression becomes
$(x+y')(x+y+z')$
Again using the identity,
$(x+A)(x+B)=x+AB$
Here,
$A=y'$ and $B=y+z'$
So,
$(x+y')(x+y+z')=x+y'(y+z')$
$=x+y'y+y'z'$
Since
$y'y=0$
So,
$=x+y'z'$
Therefore, the simplified value is
$x+y'z'$
Online Test Series, Information About Examination,
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Online Test Series, Information About Examination,
Syllabus, Notification
and More.