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algol [13]
3 years ago
10

A person using a prosthetic device can easily overheat on a hot vehicle. a. True b. False

Computers and Technology
1 answer:
Anni [7]3 years ago
4 0
Is is actually A. True

The human body regulates body temperature through the skin and sweating and being an amputee can reduce this.
You might be interested in
Program MATH_SCORES: Your math instructor gives three tests worth 50 points each. You can drop one of the test scores. The final
Simora [160]

Answer:

import java.util.Scanner;

public class num5 {

   public static void main(String[] args) {

       Scanner in = new Scanner(System.in);

       //Prompt and receive the three Scores

       int score1;

       int score2;

       int score3;

       do {

           System.out.println("Enter first Score Enter score between 1 -50");

           score1 = in.nextInt();

       } while(score1>50 || score1<0);

       do {

           System.out.println("Enter second Score.The second score must be between 1 -50");

           score2 = in.nextInt();

       } while(score2>50 || score2<0);

       do {

           System.out.println("Enter Third Score Third score must between 1 -50");

           score3 = in.nextInt();

       } while(score3>50 || score3<0);

       //Find the minimum of the three to drop

       int min, min2, max;

       if(score1<score2 && score1<score3){

           min = score1;

           min2 = score2;

           max = score3;

       }

       else if(score2 < score1 && score2<score3){

           min = score2;

           min2 = score1;

           max = score3;

       }

       else{

           min = score3;

           min2 = score1;

           max = score2;

       }

       System.out.println("your entered "+max+", "+min2+" and "+min+" the min is");

       int total = max+min2;

       System.out.println("Total of the two highest is "+total);

       //Finding the grade based on the cut-off points given

       if(total>=90){

           System.out.println("Grade is A");

       }

       else if(total>=80){

           System.out.println("Grade is B");

       }

       else if(total>=70){

           System.out.println("Grade is C");

       }

       else if(total>=60){

           System.out.println("Grade is D");

       }

       else{

           System.out.println("Grade is F");

       }

   }

}

Explanation:

  • Implemented with Java
  • Use the scanner class to receive user input
  • Use a do.....while loop to validate user input for each of the variables. A valid score must be between 0 and 50 while(score>50 || score<0);  
  • Use if and else to find the minimum of the three values and drop
  • Add the two highest numbers
  • use if/else if /else statements to print the corresponding grade
8 0
3 years ago
A personal computer can only have one operating system installed on it. true false
Pani-rosa [81]
It is true, because if you would try installing another OS after installing the first one, you would overwrite the OS which you installed before.
but in the second opinion an workstation is technically able to dual boot by installing differend operating systems on different drives. which approves that both answers can be right.
8 0
3 years ago
Purpose of this project is to increase your understanding of data, address, memory contents, and strings. You will be expected t
STALIN [3.7K]

Answer:

See explaination for the details

Explanation:

#Starting point for code/programm

main:

la $a0,newLine #Print a new line

li $v0,4

syscall

# Find the number of occurence of a string in the given sentence

la $a0,prompt1 # Prompt the user to enter the first string.

li $v0,4

syscall

li $v0, 8 # Service 8 = read input string

la $a0, fword

li $a1, 9

syscall

la $a0,prompt2 # Prompt the user to enter the second string.

li $v0,4

syscall

li $v0, 8 # Service 8 = read input string

la $a0, sword

li $a1, 9

syscall

# process first word

li $t4,0 # Intialize the couter to 0

la $t0,sstatement # Store the statement into $t0

nstart1: la $t1,fword # Store the search word into $t1

loop1: # loop1 finds the number of occurences

# of input word in the given statment

lb $t2,($t0) # Load the starting address(character) of

# sstatement into $t2

lb $t3,($t1) # Load the starting address of input word

# into $t3

beq $t3,'\n',inc_counter1

beqz $t3,inc_counter1 # If $t3 is null , exit loop and print output

beqz $t2,print_output1 # If $t2 is null , exit loop and print output

move $a0,$t2 # Convert $t2 to lower, if it is upper case

jal convert2lower

move $t2,$v0 # Store the return($v0) value into $t2

move $a0,$t3 # Convert $t3 to lower, if it is upper case

jal convert2lower

move $t3,$v0 # Store the return($v0) value into $t3

bne $t2,$t3,next_char1 # If both characters are not matched current

# character in the string, go to next character

addiu $t0,$t0,1 # otherwise, increment both indexes

addiu $t1,$t1,1

j loop1 # go to starting of the loop

next_char1:

la $t5,fword

bne $t5,$t1,nstart1

la $t1,fword # Store the input word into $t1

addiu $t0,$t0,1 # Increment the index to goto next character

j loop1 # go to starting of the loop

inc_counter1:

addi $t4,$t4,1 # Increment the frequency counter by 1

la $t1,fword # Store input word into $t1

j loop1 # go to starting of the loop

print_output1:

la $t0,fword

L1:

lb $a0,($t0)

beq $a0,'\n',exL1

jal convert2upper

move $a0,$v0

li $v0,11

syscall

addiu $t0,$t0,1

j L1

exL1:

la $a0,colon

li $v0,4

syscall

la $a0, dash

li $v0, 4

syscall

move $a0,$t4

li $v0,1

syscall # print new line

la $a0,newLine

li $v0,4

syscall

# process second word

li $t4,0 # Intialize the couter to 0

la $t0,sstatement # Store the statement into $t0

nstart2: la $t1,sword # Store the search word into $t1

loop2: # loop1 finds the number of occurences

# of input word in the given statment

lb $t2,($t0) # Load the starting address(character) of

# sstatement into $t2

lb $t3,($t1) # Load the starting address of input word

# into $t3

beq $t3,'\n',inc_counter2

beqz $t3,inc_counter2 # If $t3 is null , exit loop and print output

beqz $t2,print_output2 # If $t2 is null , exit loop and print output

move $a0,$t2 # Convert $t2 to lower, if it is upper case

jal convert2lower

move $t2,$v0 # Store the return($v0) value into $t2

move $a0,$t3 # Convert $t3 to lower, if it is upper case

jal convert2lower

move $t3,$v0 # Store the return($v0) value into $t3

bne $t2,$t3,next_char2 # If both characters are not matched current

# character in the string, go to next character

addiu $t0,$t0,1 # otherwise, increment both indexes

addiu $t1,$t1,1

j loop2 # go to starting of the loop

next_char2:

la $t5,sword

bne $t5,$t1,nstart2

la $t1,sword # Store the input word into $t1

addiu $t0,$t0,1 # Increment the index to goto next character

j loop2 # go to starting of the loop

inc_counter2:

addi $t4,$t4,1 # Increment the frequency counter by 1

la $t1,sword # Store input word into $t1

j loop2 # go to starting of the loop

print_output2:

la $t0,sword

L2:

lb $a0,($t0)

beq $a0,'\n',exL2

jal convert2upper

move $a0,$v0

li $v0,11

syscall

addiu $t0,$t0,1

j L2

exL2:

la $a0,colon

li $v0,4

syscall

la $a0, dash2

li $v0, 4

syscall

move $a0,$t4

li $v0,1

syscall

exit:

# Otherwise, end the program

li $v0, 10 # Service 10 = exit or end program

syscall

############################ subroutine - convert2lower #################################

convert2lower: # Converts a character(stored in $a0) to

# its lower case, if it is upper case

# and store the result(lower case) in $v0

move $v0,$a0

blt $a0,'A',return

bgt $a0,'Z',return

subi $v0,$a0,-32

return: jr $ra # Return the converted(lower case) character

############################## subroutine - convert2upper ##################################

convert2upper: # Converts a character(stored in $a0) to

# its upper case, if it is lower case

# and store the result(upper case) in $v0

move $v0,$a0

blt $a0,'a',return2

bgt $a0,'z',return2

addiu $v0,$a0,-32

return2: jr $ra # Return the converted(lower case) character

4 0
3 years ago
A network technician has configured a point-to-point interface on a router. Once the fiber optic cables have been run, though, t
masya89 [10]

Answer:

Wavelength Mismatch

Explanation:

Optical fiber cables are used because they have lowest attenuation(loss of signal strength).

Attenuation is directly proportional to wavelength.

According to me this issue is most likely caused by Wavelength mismatch.If the wavelength of the signal is high then there is more attenuation in that signal.

6 0
3 years ago
Consider this program segment: int newNum = 0, temp; int num = k; // k is some predefined integer value 0 while (num &gt; 10) {
Doss [256]

Answer:

All of these statements are true.

Explanation:

Since the while loop is reversing the integer number and leaving the highest order digit in the num and stores the reversed number in the newNum variable.

It skips one digit so if the num is in the range of [100,1000] it will result in a number between 10 and 100.

This loop can never go in infinite loop for any initial value of num because the loop will run as many times as the number of digits.

and if the value of the num is <=10 the while loop will never run and the value of newNum will be 0.

6 0
4 years ago
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