Devices designed primarily to be used with a company network are called

In a List of Positive Integers, Set MINIMUM to 1. For each number X in the list L, compare it to MINIMUM. If X is smaller, set M

Dvinal [7]

Answer:

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

Explanation:

The given problem scenario is:

In a List of Positive Integers, Set MINIMUM to 1. For each number X in the list L, compare it to MINIMUM. If X is smaller, set MINIMUM to X.

The answer to question A is: this algorithm will not run because:

If you a list of a positive number and you have already set it MINIMUM to 1, the comparison result with MINIMUM will not obtain. For example, lets suppose a list of numbers L. L={1,2,3,1,1,5,4,3,2}. You have already set the Minimum to 1. When you compare each number X in the list, if X >MINIMUM, then set MINIMUM to X.

This will not work, because there are positive numbers in the list and the minimum positive number is 1. So the algorithm, will not execute the if-part.

In short, the pseudo-code of this problem is given below:

L={list of positive number e.g. 1,2,3,.......Xn}

MINIMUM=1

foreach ( X in L)

{

(if X<MINIMUM)

{

MINIMUM=X

}

}

The answer to question B is X will not be replaced because X replaced with MINIMUM only when if-part of this algorithm get executed. However, it is noted that X will not replace any value, if if-part will get executed, the Variable MINIMUM will get replaced with any value.

4 0

3 years ago

What will the following code display? #include using namespace std; void doSomething(int); int main() { int x{2}; cout <<

Leno4ka [110]

Answer:

The code will display the following on the screen:-

2

0

2

Explanation:

The output is like this because x{2} assigns the value 2 to the integer x.First the value of x is printed on the screen that is 2 and that newline is used to go to the next line.After that function is called doSomething with x passed as the argument. doSomething function assigns the value zero to the variable passed and prints it on the screen.So the next line is 0.Since x is passed by value so no changes will appear in the original argument.Hence the last line printed is 2.

8 0

3 years ago

The address of the last cell of a memory RAM is 3FFFFh.a total capacity of the main memory and 4M bits and the data bus can tran

marysya [2.9K]

Hey there!:

Given the address of the last cell of a memory RAM is 3FFFFh then :

There are 8 bits in a byte

1024 bytes in a kilobyte

so 8*1024= 8192

so 8192 bits in a kb

32*8192 = 262144

there are 262144 bits in 32KB .

a) 32 bit address registers must match 2^32 byte = 4 GB of physical memory.

However, with 32 Bit you can also address more than 4 GB, like Physical Address Extension (PAE) does.

The other way is that you can access 4 GB with less than 32 bit address register.

____________________________________________________

b) Main Memory = 4M × 32Kbits, RAM chips = 32K × 4bit.

For this memory we require 4 × 2 = 8 RAM chips.

Each chip requires 18 address bits (ie. 218 = 256K).

And 1M × 8 bits requires 20 address bits (ie. 220 = 1M )

_____________________________________________________

c) The size of the storage cells is known as the word size for the computer.

In some computers, the word size is one byte while in other computers the word size is two, four, or even eight bytes. In our 4M main memory the size of cell is 4*2^20 bytes.

Each storage cell in main memory has a particular address which the computer can use for storing or retrieving data.

______________________________________________________

d) The size of MBR is the 4M bits of main memory.

_____________________________________________________

Hope this helps!

6 0

3 years ago

Explain why you will do these prarticular things when driving

VashaNatasha [74]

What things?????? Are u talking about

4 0

4 years ago

X = 10 y = 20 x > y print("if statement") print("else statement")

Eduardwww [97]

I dont really understand what you want to do but im gonna answer it as if I understood it.

If x is 10 and y is 20 then y is larger than x

8 0

3 years ago