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wariber [46]
3 years ago
11

The process of hazard recognition, evaluation, and control is the foundation of an effective___program.

Engineering
1 answer:
Alona [7]3 years ago
7 0
Answer: A. Safety program
You might be interested in
1. The system must be able to manage multiple students (max of 15) and their grades for assignments from three different assignm
mr_godi [17]

Answer:

#include<iostream>

#include<string.h>

using namespace std;

class gradebook;

class student

{

friend class gradebook;

char name[20]; // std name

int homework_grades[6], quiz_grades[6], exam_grades[4]; //grades array

float homework_avg, quiz_avg, exam_avg, final_avg;

};

class gradebook

{

student list[16];

public:

char gb_name[20]; // gradebook name

int max, total;

// Constructor

gradebook()

{

strcpy(gb_name,"noname");

max = 5;

total = 0;

for(int i=1;i<=15;i++)

{

strcpy(list[i].name,"noname");

for(int j=1;j<=5;j++)

{

list[i].homework_grades[j] = -1;

list[i].quiz_grades[j] = -1;

}

list[i].exam_grades[1] = list[i].exam_grades[2] = list[i].exam_grades[3] = -1;

}

}

// Member functions

void display_std_name();

void display_grades(int, int);

int add_std(char *new_name);

int remove(int);

//int ch_grade(int, int);

};

// Define all functions

// Remove a student from list and adjust remaining

int gradebook::remove(int std_id)

{

if(total==0 || std_id>total || std_id<=0)

return 0;

if(std_id==total)

{

cout<<"\n“"<<list[std_id].name<<"” has been successfully removed! (New class size: 2)";

strcpy(list[total--].name,"noname");

return 1;

}

cout<<"\n“"<<list[std_id].name<<"” has been successfully removed! (New class size: 2)";

for(int i=std_id; i<total; i++)

{

list[i]=list[i+1];

}

total--;

return 1;

}

// Add student to gradebook

int gradebook::add_std(char *new_name)

{

if(total<15)

{

strcpy(list[++total].name, new_name);

return 1;

}

return 0;

}

// Display student name list

void gradebook::display_std_name()

{

for(int i=1;i<=total;i++)

{

cout<<"\n"<<i<<" "<<list[i].name;

}

}

/* Display grades of students

// (grade_id, student_number_in_list)

homework(1), quiz(2), exam(3)

*/

void gradebook::display_grades(int g_id, int std_id)

{

int i;

if(std_id<=total)

{

if(g_id==1)

{

for(i=1;i<6;i++)

{

cout<<"\n"<<i<<". ";

if(list[std_id].homework_grades[i]==-1) cout<<"<ungraded>";

}

}

else if(g_id==2)

{

for(i=1;i<6;i++)

{

cout<<"\n"<<i<<". ";

if(list[std_id].quiz_grades[i]==-1) cout<<"<ungraded>";

}

}

else if(g_id==3)

{

for(i=1;i<4;i++)

{

cout<<"\n"<<i<<". ";

if(list[std_id].exam_grades[i]==-1) cout<<"<ungraded>";

}

}

}

else

cout<<"\nNo such std exists";

}

/****************************** MAIN ****************************/

int main()

{

int main_choice, grade_choice, std_id, new_grade;

char g_name[20], s_name[20];

float a,b,c;

gradebook g;

do

{

cout<<"\n\n-=| MAIN MENU |=-";

cout<<"\n1. Add a student\n2. Remove a student\n3. Change a student’s grade\n4. Change group weights\n5. Change the gradebook name\n6. Display class averages\n7. Display full report\n0. QUIT\n";

cout<<"\nEnter an action: ";

cin>>main_choice;

switch(main_choice)

{

case 0:

break;

case 1:

cout<<"\n-=| ADDING STUDENT |=-";

cout<<"\nPlease enter the student’s name: ";

cin>>s_name;

if(g.add_std(s_name))

cout<<"\n"<<s_name<<" was successfully added to the gradebook!";

else

cout<<"\nStudents cannot be added because the gradebook is full!";

break;

case 2:

cout<<"\n-=| REMOVING STUDENT |=-";

if(g.total!=0)

{

g.display_std_name();

cout<<"\n Enter student to remove: ";

cin>>std_id;

if(!g.remove(std_id))

cout<<"\nCannot remove students";

}

else

cout<<"\nStudents cannot be added because the gradebook is full!";

break;

case 3:

cout<<"\n-=| CHANGING GRADE |=-\n1. Change a homework grade\n2. Change a quiz grade\n3. Change an exam grade";

cout<<"\nWhat type of grade would you like to change: ";

cin>>grade_choice;

g.display_std_name();

cout<<"\nWhich student’s grade would you like to change? ";

cin>>std_id;

//ch_grade(grade_choice, std_id);

g.display_grades(grade_choice, std_id);

break;

/*

case 4:

cout<<"\n-=| CHANGING WEIGHTS |=-";

while(1)

{

cout<<"\nEnter the weights, separated by spaces, in the order of homework, quizzes, and exams (total must add up to 1.0):";

cin>>a>>b>>c;

if(a+b+c==1.00)

break;

else

cout<<"\n Weights do not add up to 1.0, try again...";

}

break;

case 5:

cout<<"\n-=| CHANGING NAME |=-\nPlease enter the new name for the gradebook: ";

cin>>g_name;

cout<<"\n Gradebook name changed from “ ” to "<<g_name;

break;

case 6:

cout<<"\n-=| CLASS AVERAGES |=-";

cout<<"\nHomework average of class: ";

cout<<"\nQuiz average of class: ";

cout<<"\nExam average of class: ";

cout<<"\nFinal average of class: ";

break;

case 7:

break;

*/

default:

cout<<"\n\nInvalid input\n\n";

}

}while(main_choice!=0);

return 0;

}

Explanation:

6 0
3 years ago
From the following numbered list of characteristics, decide which pertain to (a) precipitation hardening, and which are displaye
tia_tia [17]

Answer:

(a) Precipitation hardening - 1, 2, 4

(b) Dispersion strengthening - 1, 3, 5

Explanation:

The correct options for each are shown as follows:

Precipitation hardening

From the first statement; Dislocation movement is limited by precipitated particles. This resulted in an expansion in hardness and rigidity. Precipitates particles are separated out from the framework after heat treatment.

The aging process occurs in the second statement; because it speaks volumes on how heated solutions are treated with alloys above raised elevated temperature. As such when aging increases, there exists a decrease in the hardness of the alloy.

Also, for the third option for precipitation hardening; This cycle includes the application of heat the alloy (amalgam) to a raised temperature, maintaining such temperature for an extended period of time. This temperature relies upon alloying components. e.g. Heating of steel underneath eutectic temperature. Subsequent to heating, the alloy is extinguished and immersed in water.

Dispersion strengthening

Here: The effect of hearting is not significant to the hardness of alloys hardening by the method in statement 3.

In statement 5: The process only involves the dispersion of particles and not the application of heat.

8 0
3 years ago
What is the essence of the Central park and it impact in New York?​
Mars2501 [29]
Central Park’s significance was its influence throughout nationwide park architecture. It’s also an iconic landmark of New York and attracts millions of people each year
3 0
3 years ago
25 gallons of an incompressible liquid exert a force of 70 lbf at the earth’s surface. What force in lbf would 6 gallons of this
jekas [21]

Answer:

froce by 6 gallon liquid on moon surface is  2.86 lbf

Explanation:

given data:

at earth surface

volume of an incompressible liquid = Ve = 25 gallons

force by liquid = 70 lbf

on moon

volume of  liquid = Vm = 6 gallons

gravitational acceleration on moon is am = 5.51 ft/s2

Due to incompressibility , the density remain constant.

mass of liquid on surface of earth= \frac{ force}{ acceleration}

mass = \frac{70lbf}{32.2 ft/s2}

mass = 2.173 pound

density \rho = \frac{mass}{volume}

                  = \frac{2.173}{25} = 0.0869 pound/ gallon

froce by 6 gallon liquid on moon surface is

Fm = mass * acceleration

      = density* volume * am

      = 0.0869 *6* 5.51

      = 2.86 lbf

5 0
3 years ago
1. One of these is NOT a type of pneumatic tool. Which one?
Serggg [28]

Answer:Circular

Explanation:

It’s the only thing not list under pneumatic tools‍♂️

5 0
3 years ago
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