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Maru [420]
1 year ago
7

In C++ please (read the image below for instructions)

Computers and Technology
1 answer:
FromTheMoon [43]1 year ago
8 0

Using the knowledge in computational language in C code  it is possible to write a code that  print the month name along with their rainfall inches. Then we have to calculate the total rainfall.

<h3>Writting the code:</h3>

<em>#include <bits/stdc++.h></em>

<em>using </em><em>namespace </em><em>std;</em>

<em />

<em>// function to print the rainfall of each month</em>

<em>void rainfallStatistics(string month[], double rainfall[]){</em>

<em>    </em>

<em>    double sum = 0; // variable to store the total sum of rainfall</em>

<em>    string maxMonth = month[0]; // variable to return month with maximum rainfall</em>

<em>    double </em><em>maxRainfall </em><em>= INT_MIN; // variable to store the maximum rainfall</em>

<em>    </em>

<em>    // running the loop till the last month</em>

<em>    for(int i=0; i<12; i++){</em>

<em>        </em>

<em>        // printing the month and rainfall</em>

<em>        cout<< month[i] << "           " << fixed << setprecision(2) << rainfall[i] << " inches" << endl;</em>

<em>        </em>

<em>        // storing the sum</em>

<em>        sum = sum + rainfall[i];</em>

<em>        </em>

<em>        // checking the month with highest rainfall</em>

<em>        if(rainfall[i] > </em><em>maxRainfall</em><em>){</em>

<em>            </em>

<em>            maxRainfall = rainfall[i];</em>

<em>            maxMonth = month[i];</em>

<em>        }</em>

<em>    }</em>

<em>    </em>

<em>    // calculating the average of the rainfall</em>

<em>    double average = sum / double(12);</em>

<em>    </em>

<em>    cout << endl;</em>

<em>    cout << "Total for the year " << sum << " inches" << endl;</em>

<em>    cout << "Average rainfall per month " << average << " inches" << endl;</em>

<em>    cout << "Month with the greatest rainfall was " << maxMonth << " with rainfall of " << </em><em>maxRainfall </em><em><< " inches" << endl;</em>

<em>    </em>

<em>}</em>

<em>// driver program</em>

<em>int main()</em>

<em>{</em>

<em>    // array to store the nonth name</em>

<em>    string month[12] = {"Jan", "Feb", "Mar", "Apr", "May", "Jun", "Jul", "Aug", "Sep", "Oct", "Nov", "Dec"};</em>

<em>    </em>

<em>    // array to store the rainfall</em>

<em>    double rainfall1[12] = { 4.50, 4.25, 3.26, 1.35, 0.80, 0.20, 0.10, 0.00, 0.40, 1.20, 2.96, 4.71 };</em>

<em>    double rainfall2[12] = { 3.50, 1.25, 7.26, 8.35, 2.80, 0.00, 1.10, 4.00, 9.40, 2.20, 3.96, 4.71 };</em>

<em>    </em>

<em>    cout << "Hayward Statistics" << endl;</em>

<em>    cout << endl;</em>

<em>    cout<< "Month" << "         " << "Rainfall" << endl;</em>

<em>    </em>

<em>    // calling the function for Hayward</em>

<em>    </em><em>rainfallStatistics</em><em>(month, rainfall1);</em>

<em>    cout << endl;</em>

<em>    cout << endl;</em>

<em>    </em>

<em>    cout << "Houston Statistics" << endl;</em>

<em>    cout << endl;</em>

<em>    cout<< "Month" << "         " << "Rainfall" << endl;</em>

<em>    </em>

<em>    // calling the function for Houston</em>

<em>    rainfallStatistics(month, rainfall2);</em>

<em>    </em>

<em>    return 0;</em>

<em>}</em>

See more about c code at brainly.com/question/19705654

#SPJ1

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Proved

Explanation:

Given

X = (a.\bar b)+(\bar a.b)

(a + b)\ .  \frac{}{a.b}

Required

Find out why they represent the same

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In this question, I will simplify (a + b)\ .  \frac{}{a.b}

Apply de morgan's law

(a + b)\ .  \frac{}{a.b} = (a + b) . (\bar a + \bar b)

Apply distribution property

(a + b)\ .  \frac{}{a.b} = a.\bar a + a.\bar b + \bar a . b + b . \bar b

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(a + b)\ .  \frac{}{a.b}  = a.\bar b + \bar a.b

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(a + b)\ .  \frac{}{a.b}  = (a.\bar b)+(\bar a.b)

From the given parameters:

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This implies that:

(a + b)\ .  \frac{}{a.b} when simplified is X or (a.\bar b)+(\bar a.b)

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