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3 years ago

Write a program to input 6 numbers. After each number is input, print the biggest of the numbers entered so far.

Engineering

1 answer:

likoan [24]3 years ago

4 0

Answer:

P<u>rogram:</u>

# Enter Numbers #

number1 = int(input("Enter number: " ))

print("Largest: " + string(number1))

#for num 2 #

number2 = int(input("Enter a number: "))

if number2 > number1:

print("Largest: " + string(number2))

else:

print("Largest: " + string(num1))

#for num 3 #

number3 = int(input("Enter a number: "))

print("Largest: " + string(max(number1, number2, number3)))

#for num 4 #

number4 = int(input("Enter a number: "))

print("Largest: " + string(max(number1, number2, number3, number4)))

#for num 5 #

number5 = int(input("Enter a number: "))

print("Largest: " + string(max(number1, number2, number3, number4, number5)))

#for num 6 #

number6 = int(input("Enter a number: "))

print("Largest: " + string(max(number1, number2, number3, number4, number5, number6)))

# END #

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Solid Isomorphous alloys strength

Sphinxa [80]

Answer:

Explanation:

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3 0

3 years ago

Modify the Rainfall Statistics program you wrote for Programming Challenge 2 of Chapter 7 . The program should display a list of

rjkz [21]

Answer:

#include<iostream>

#include <iomanip>

using namespace std;

const int NUM_MONTHS = 12;

double getTotal(double [], int);

double getAverage(double [], int);

double getLargest(double [], int, int &);

double getSmallest(double [], int, int &);

double getTotal(int rainFall,double NUM_MONTHS[])

{

double total = 0;

for (int count = 0; count < NUM_MONTH; count++)

total += NUM_MONTH[count];

return total;

}

double getAverage(int rainFall,double NUM_MONTH[])

{getTotal(rainFall,NUM_MONTH)

average= total/NUM_MONTHS;

return average;

}

double getHighest(int rainFall, double NUM_MONTHS[]) //I left out the subScript peice as I was not sure how to procede with that;

{

double largest;

largest = NUM_MONTHS[0];

for ( int month = 1; month <= NUM_MONTHS; month++ ){

if ( values[month] > largest ){

largest = values[month];

return largest;

}

double getSmallest(int rainFall, double NUM_MONTHS[])

{

double smallest;

smallest = NUM_MONTHS[0];

for ( int month = 1; month <= NUM_MONTHS; month){

if ( values[month] < smallest ){

smallest = values[month];

return smallest;

}

int main()

{

double rainFall[NUM_MONTHS];

for (int month = 0; month < NUM_MONTHS; month++)

{

cout << "Enter the rainfall (in inches) for month #";

cout << (month + 1) << ": ";

cin >> rainFall[month];

while (rainFall[month] < 0)

{

cout << "Rainfall must be 0 or more.\n"

<< "Please re-enter: ";

cin >> rainFall[month];

}

cout << fixed << showpoint << setprecision(2) << endl;

cout << "The total rainfall for the year is ";

cout << getTotal(rainFall, NUM_MONTHS)

<< " inches." << endl;

cout << "The average rainfall for the year is ";

cout << getAverage(rainFall, NUM_MONTHS)

<< " inches." << endl;

int subScript;

cout << "The largest amount of rainfall was ";

cout << getLargest(rainFall, NUM_MONTHS, subScript)

<< " inches in month ";

cout << (subScript + 1) << "." << endl;

cout << "The smallest amount of rainfall was ";

cout << getSmallest(rainFall, NUM_MONTHS, subScript)

<< " inches in month ";

cout << (subScript + 1) << "." << endl << endl;

return 0;

}

8 0

4 years ago

Anne works with other engineers studying physical processes that involve the flow of particles. Which field of engineering would

slega [8]

Answer:

Option A

Chemical engineering

Explanation:

Chemical engineering mainly encompass the study of behavior of different particles such as petroleum, water, drugs and other products. When Anne is involved in a study with engineers who study flow of particles, the flow, viscosity and other properties are among the behavior that chemical engineers are involved in.

3 0

4 years ago

Gtjffs

grandymaker [24]

the required documents is 3000

4 0

3 years ago

How many electrons move past a fixed reference point every t = 2.55 ps if the current is i = 7.3 μA ? Express your answer as an

iris [78.8K]

Answer:

116.3 electrons

Explanation:

Data provided in the question:

Time, t = 2.55 ps = 2.55 × 10⁻¹² s

Current, i = 7.3 μA = 7.3 × 10⁻⁶ A

Now,

we know,

Charge, Q = it

thus,

Q = (7.3 × 10⁻⁶) × (2.55 × 10⁻¹²)

Q = 18.615 × 10⁻¹⁸ C

Also,

We know

Charge of 1 electron, q = 1.6 × 10⁻¹⁹ C

Therefore,

Number of electrons past a fixed point = Q ÷ q

= [ 18.615 × 10⁻¹⁸ ] ÷ [ 1.6 × 10⁻¹⁹ ]

= 116.3 electrons

4 0

3 years ago