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jenyasd209 [6]
3 years ago
15

The word count of the active document is typically displayed on the ________.

Computers and Technology
1 answer:
romanna [79]3 years ago
4 0

Click the Word Count in the Proofing Group. Maybe the name is the Bottom ribbon tab

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"how do we store information in long term memory?"
Andru [333]

Answer:

By saving it in the hard drive.

Hope I helped!

8 0
3 years ago
A signal travels from point A to point B. At point A, the signal power is 100 W. At point B, the power is 90 W. What is the atte
11Alexandr11 [23.1K]

Answer:

Attenuation = 0.458\ db

Explanation:

Given

Power at point A = 100W

Power at point B = 90W

Required

Determine the attenuation in decibels

Attenuation is calculated using the following formula

Attenuation = 10Log_{10}\frac{P_s}{P_d}

Where P_s = Power\ Input and P_d  = Power\ output

P_s = 100W

P_d = 90W

Substitute these values in the given formula

Attenuation = 10Log_{10}\frac{P_s}{P_d}

Attenuation = 10Log_{10}\frac{100}{90}

Attenuation = 10 * 0.04575749056

Attenuation = 0.4575749056

Attenuation = 0.458\ db <em>(Approximated)</em>

7 0
3 years ago
Create a program to deteate a program to determine whether a user-specified altitude [meters] is in the troposphere, lower strat
allsm [11]

Answer:

#include <iostream>

using namespace std;

int main()

{

   float altitude;

   cout<<"Enter alttitude in meter";

cin >>altitude;

if(altitude<0 || altitude > 50000)

{

   cout<<"Invalid value entered";

   return 0;

}

else{

   if(altitude<= 10000)

   {cout <<"In troposphere "<<endl;}

   

   else if(altitude>10000 && altitude<=30000)

   {cout <<"In lower stratosphere"<<endl;}

   

   else if(altitude>30000 && altitude<=50000)

   {cout <<"In upper stratosphere"<<endl;}

   

}

   return 0;

}

Explanation:

Define a float type variable. Ask user to enter altitude in meters. Store value to altitude variable.

Now check if value entered is positive and less than 5000,if its not in the range of 0-50,000 display a termination message and exit else check if it's in less than 10000, in between 10000 and 30000 or in between 30000 and 50000.

10,000 is above sea level is troposphere.

10,000-30,000 is lower stratosphere.

30,000-50,000 is upper stratosphere.

8 0
4 years ago
Convert the following numbers. (Please show all steps; no marks will be awarded if no steps are shown) [1.5 x 4 = 6 marks]
katen-ka-za [31]

Answer:

i. 10210212

ii. 100 101 001 in BASE 2

iii. 46.4631

iv. 12.453125

Explanation:

i. Converting to base 10 we get

=  10 x 162 + 13 x 161 + 9 x 160 = 2777

Converting to base 10 we get  =

2777/3 = 925 with remainder 2

925/3 = 308 with remainder 1

308/3 = 102 with remainder 2

102/3 = 34 with remainder 0

34/3 = 11 with remainder 1

11/3 = 3 with remainder 2

3/3 = 1 with remainder 0

1/3 = 0 with remainder 1

Hence the Answer is 10210212

ii. = Octal is taken in pair of 3 bits , Writing binary of each number in 3 bits we get  = 100 101 001 in BASE 2

iii. = 1 x 52 + 2 x 51 + 3 x 50 + 3 x 5-1   = 38.6 in decimal

Converting to Octal we get

38 /8 = 4 with remainder 6

4 /8 = 0 with remainder 4

.6 x 8 =  4.8

.8 x 8 = 6.4

46.4631

iv. = 1x 81 + 4 x 80 + 3 x 8-1 + 5 x 8-2   = 12.453125

Check here for Free courses about IT and Management www.eacademy.lk

4 0
3 years ago
Write the half function. A function call and the functionprototype
EastWind [94]

Answer:

C code for half()

#include<stdio.h>

void half(float *pv);

int main()

{

float value=5.0;  //value is initialized  

printf ("Value before half: %4.1f\n", value); // Prints 5.0

half(&value);  // the function call takes the address of the variable.

printf("Value after half: %4.1f\n", value); // Prints 2.5

}

void half(float *pv) //In function definition pointer pv will hold the address of variable passed.

{

*pv=*pv/2;  //pointer value is accessed through * operator.

}

  • This method is called call-by-reference method.
  • Here when we call a function, we pass the address of the variable instead of passing the value of the variable.
  • The address of “value” is passed from the “half” function within main(), then in called “half” function we store the address in float pointer ‘pv.’ Now inside the half(),  we can manipulate the value pointed by pointer ‘pv’. That will reflect in the main().
  • Inside half() we write *pv=*pv/2, which means the value of variable pointed by ‘pv’ will be the half of its value, so after returning from half function value of variable “value” inside main will be 2.5.

Output:

Output is given as image.

3 0
3 years ago
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