Does the sticker list an automatic or manual transmission?

In an SLR camera photogaphers look from the viewfinder through which of the following?

Fantom [35]

The answer is A mirror the lens

8 0

3 years ago

Read 2 more answers

2.8 Code Practice: Question 1

-BARSIC- [3]

Answer:

The program to this question can be given as:

Program:

#include <stdio.h> //define header file

int main() //define main method.

{

int a1,b1,c1; //define variable

printf("Enter three numbers:\n"); //message

scanf("%d",&a1); //input numbers.

scanf("%d",&b1);//input numbers.

scanf("%d",&c1);//input numbers.

if (a1>b1) //check condition

{

//check condition

if(a1>c1)//inner if block

{

printf("The largest number: %d", a1); //message

}

else //inner else block

{

printf("The largest number: %d", c1);//message

}

else //else block

{

//check condition

if(b1>c1) //inner if block.

{

printf("The largest number: %d", b1);//message

}

else //else block

{

printf("The largest number: %d",c1);//message

}

return 0; //return 0.

}

Output:

Enter three numbers:

22

44

11

The largest number: 44

Explanation:

The description of the above program can be given as:

In the C language program firstly we define the header file. Then we define the main function. In the main function, we define three variables that are "a1, b1 and c1". In this variable, we take input from the user.
Then we define the conditional statements. In the if block we check if the value of a1 is greater then the value of b1 if this condition is true so we use another condition that if the value of a1 is greater then the value of c1. so, it will print the largest value. If the above condition is not true then it will go to else section.
In the else part we check the condition that if the value of b1 is greater then the value of c1. if it is true it will print the largest number value.

4 0

3 years ago

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Problem 2 Write a method that (a) returns an int[] , (b) takes in two sorted arrays (in descending order) and (c) returns an int

Galina-37 [17]

Answer:

Following are the Program in the Java Programming Language.

//define class

public class Main

{

//define function to merge and sort array

public static int[] sort_Array(int[] a, int[] b)

{

//set integer variable and initialize values

int x = 0, y = 0, k = 0, num = 0;

int n = a.length;

int n1 = b.length;

//set integer type array

int[] ar = new int[n+n1];

//set while loop

while (x<n && y <n1)

{

num++;

//set if conditional statement

if (a[x] > b[y])

ar[k++] = a[x++];

else

ar[k++] = b[y++];

num++;

}

//set the while loop

while (x < n)

{

num++;

ar[k++] = a[x++];

}

//set the while loop

while (y < n1)

{

num++;

ar[k++] = b[y++];

}

//print the total compares done

System.out.println("Number of compares done: " + n);

//return array

return ar;

}

//define main method

public static void main(String[] args)

{

//declaration and the initialization of array

int[] a = new int[]{18,16, 15, 6, 2};

int[] b = new int[]{13, 12, 11, 7, 1};

int[] ar;

//call function

ar = sort_Array(a, b);

//print message

System.out.println("\nThe resultant array is: \n");

//set for loop to print array

for(int x=0; x<ar.length; x++)

{

System.out.print(ar[x] + " ");

}

<u>Output</u>:

Number of compares done: 5

The resultant array is:

18 16 15 13 12 11 7 6 2 1

Explanation:

Here, we define a class "Main" inside the class, define a function "sort_Array()"and pass two integer type array argument "a" and "b".

Set four integer type variables "x", "y", "k", and "num" and initialize them to 0.
Set two integer type array "n", "n1" and initialize them the length of the array then, set integer type array variable "ar".
Set the while loop then, set if conditional statement then, agin set while loop to sort array in descending order.
Then, print the number of compares and return the value of the variable "ar".

Finally, define the main method inside it, we set two integer type array variable and initialize elements in them then, set the for loop to print the sorted elements of the array.

4 0

3 years ago

Determine the number of character comparisons made by the brute-force algorithm in searching for the pattern GANDHI in the text

leonid [27]

Answer:

Total number of character comparison = 43

Explanation:

Using the Brute force algorithm

The string of n characters is known as text, and the string of m characters is known as the pattern.

From the given information:

The text (n)=THERE_IS_MORE_TO_LIFE_THAN_INCREASING_ITS_SPEED

The pattern (m) = GANDHI

The total no of characters that we have in the text = 47

The total number of characters in pattern = 6

For a brute force algorithm;

Since; the first character of the pattern does not exist in the text, then the number of trials made can be attempted can be expressed as = n – m + 1

= 47 – 6 + 1

= 47 – 5

= 42

Thus; the algorithm will attempt the trial 42 times.

Now, for loop in the algorithm to run 42 times, the G in the pattern will have to align against the for T in the text, and in the last case, it will be aligned against the last space.

On each attempted trial, the algorithm will make one unsuccessful comparison.

However, at the trial at which the G in the pattern Is aligned with the G in the text, there will be two successful comparisons.

Hence, we can calculate the total number of character comparison as follows:

Total number of character comparison = $\mathbf{\bigg ( ( 42 - (no. \ of \ failed \ comparison) ) \times 1 + (1 \times ( Two \ successful \ comparisons) ) \bigg ) }$

Total number of character comparison = ( (( 42 – 1) × 1 ) + ( 1 × 2) )

Total number of character comparison = 41 + 2

Total number of character comparison = 43

3 0

3 years ago

1) Create a recursive program that prompts the user for a nonnegative integer N and outputs. a) Sum [N] b) BiPower [N] c) TimesF

Georgia [21]

Answer:

//Code is created using java

import java.util.*;

// returns the sum

public int sum(int N)

{

if(N==1)

return (1);

else

return N+sum(N-1);

}

// code to return the Bipower ouput

public int BiPower(int N)

{

if(N==1)

return (2);

else

return 2*BiPower(N-1);

}

// Code to return TimesFive output

public int TimesFive(int N)

{

if(N==1)

return 5;

else

return 5 + timesFive(N-1);

}

public static void main(String args[])

{

//Prompts the user to enter a nonnegative integer

int N = Integer.parseInt.(console.readLine("Enter a nonnegative integer: "));

//Outputs the sum, Bipower and TimesFive

System.out.println(sum(n));

System.out.println(BiPower(n));

System.out.println(TimesFive(n));

}

4 0

3 years ago