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2 years ago

40 points for this question

Computers and Technology

2 answers:

julsineya [31]2 years ago

6 0

Answer:

-promoting the common good

-helping and leading others

targeting potential cyber criminals

Explanation:

got it correct on edge!

qaws [65]2 years ago

3 0

Answer:

I think the answers are

<h3>the second one </h3><h3>the third one </h3><h3>the fourth one </h3>

hope it helps!

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1. How many bits would you need to address a 2M × 32 memory if:

Dominik [7]

Answer:

a) 23 b) 21
a) 43 b) 42
a) 0 b) 0

Explanation:

<u>1) How many bits is needed to address a 2M * 32 memory </u>

2M = 2^1*2^20, while item =32 bit long word

hence ; L = 2^21 ; w = 32

a) when the memory is byte addressable

w = 8; L = ( 2M * 32 ) / 8 = 2M * 4

hence number of bits = log2(2M * 4)= log2 ( 2 * 2^20 * 2^2 ) = 23 bits

b) when the memory is word addressable

W = 32 ; L = ( 2M * 32 )/ 32 = 2M

hence the number of bits = log2 ( 2M ) = Log2 (2 * 2^20 ) = 21 bits

<u>2) How many bits are required to address a 4M × 16 main memory</u>

4M = 4^1*4^20 while item = 16 bit long word

hence L ( length ) = 4^21 ; w = 16

a) when the memory is byte addressable

w = 8 ; L = ( 4M * 16 ) / 8 = 4M * 2

hence number of bits = log 2 ( 4M * 2 ) = log 2 ( 4^1*4^20*2^1 ) ≈ 43 bits

b) when the memory is word addressable

w = 16 ; L = ( 4M * 16 ) / 16 = 4M

hence number of bits = log 2 ( 4M ) = log2 ( 4^1*4^20 ) ≈ 42 bits

<u>3) How many bits are required to address a 1M * 8 main memory </u>

1M = 1^1 * 1^20 , item = 8

L = 1^21 ; w = 8

a) when the memory is byte addressable

w = 8 ; L = ( 1 M * 8 ) / 8 = 1M

hence number of bits = log 2 ( 1M ) = log2 ( 1^1 * 1^20 ) = 0 bit

b) when memory is word addressable

w = 8 ; L = ( 1 M * 8 ) / 8 = 1M

number of bits = 0

5 0

3 years ago

Write the definitions for three function named max. Each receives two parameters, of the same type, and returns the larger of th

GenaCL600 [577]

Answer:

The method definition to this question can be given as:

Method definition:

double max(double x, double y) //define method with double parameter

{

if (x>=y) //check condition.

return x; //return value

else

return y; //return value

}

double max(int x, int y) //define method with integer parameter

{

if (x>=y) //check condition

return x; //return value

else

return y; //return value

}

double max(char x, char y) //define method with char parameter

{

if (x>=y) //check condition

return x; //return value

else

return y; //return value

}

Explanation:

The above method definition can be described as below:

In the first method definition first, we define a method that is "max()". In this method we pass two variables as a parameter that is "x and y" and the datatype of this is double. Then we use a conditional statement. In the if block we check if variable x is greater then equal to y then it will return x else it will return y.
In the second method definition, we define a method that is same as the first method name but in this method, we pass two integer variable that is "x and y". Then we use a conditional statement. In the if block we check if variable x is greater then equal to y then it will return x else it will return y.
In the third method definition, we define a method that is same as the first and second method name but in this method, we pass two char variable that is "x and y". Then we use a conditional statement. In the if block we check if variable x is greater then equal to y then it will return x else it will return y.

8 0

3 years ago

Write a program to draw a text-based graph of a mathematical function f(x)

Makovka662 [10]

I have a very good example of the program you need written on Python. You can use this (sorry for bad tabulation):
import math import math def main(): function = input("Enter a function f(x):\n") x = 0 y = 0 for rows in range(10,-11,-1): for col in range(-10,11,1): x=col roundfx = round(eval(function)) if roundfx == rows: print("o", end="") if rows==0 and col==0 and not rows == roundfx: print("+", end="") if col == 0 and not rows == 0 and not rows == roundfx: print("|", end="") if rows==0 and not col==0 and not rows == roundfx: print("-", end="") else: if not rows == 0: if not col == 0: if not rows == roundfx: print(" ", end="") print() main()

4 0

3 years ago

Which are examples of copyrighted online materials? Check all that apply.

Stella [2.4K]

The answers are 1, 3, and 5.

7 0

3 years ago

Read 2 more answers

Again, suppose that n and r denote integer values where r > 0 and n ≥ 0. What are two "simple" values, say lowEnough and tooH

enyata [817]

Answer:

Answer explained

Explanation:

From the previous question we know that while searching for n^(1/r) we don't have to look for guesses less than 0 and greater than n. Because for less than 0 it will be an imaginary number and for rth root of a non negative number can never be greater than itself. Hence lowEnough = 0 and tooHigh = n.

we need to find 5th root of 47226. The computation of root is costlier than computing power of a number. Therefore, we will look for a number whose 5th power is 47226. lowEnough = 0 and tooHigh = 47226 + 1. Question that should be asked on each step would be "Is 5th power of number < 47227?" we will stop when we find a number whose 5th power is 47226.

6 0

3 years ago