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never [62]
1 year ago
11

From the set $\{1, 2, 3, \dots, 20\},$ ten numbers are chosen at random, forming a subset. Let $M$ be the largest element among

the ten numbers. Find the expected value of $M.$
Business
1 answer:
Marysya12 [62]1 year ago
6 0

The largest element can be as small as 10, which happens when the subset is {1, 2, ..., 10}.  The probability of choosing this subset is (1/2)^10 = 1/1024.  (Every element from 1 to 10 can either be in the subset, or not.)

The largest element can also be 11.  All the numbers in the subset must be from 1 to 10, and we must choose 1 to leave out, so the probability that the largest element is 11 is C(10,1)*1/1024.

The largest element can also be 12.  All the numbers in the subset must be from 1 to 11, and we must choose 2 to leave out, so the probability that the largest element is 12 is C(11,2)*1/1024.

We can do the other cases similarly:

Largest element is 13 -> C(12,3)*1/1024

Largest element is 14 -> C(13,4)*1/1024

Largest element is 15 -> C(14,5)*1/1024

Largest element is 16 -> C(15,6)*1/1024

Largest element is 17 -> C(16,7)*1/1024

Largest element is 18 -> C(17,8)*1/1024

Largest element is 19 -> C(18,9)*1/1024

Largest element is 20 -> C(19,10)*1/1024

Adding these up, we get (1 + C(10,1) + C(11,2) + ... + C(19,10))*1/1024.  Since 1 = C(9,0), we also get (C(9,0) + C(10,1) + C(11,2) + ... + C(19,10))*1/1024.

By the Hockey Stick Identity, C(9,0) + C(10,1) + C(11,2) + ... + C(19,10) = C(20,10), so the expected value of the largest element is 1/11*C(20,10)*1/1024 = 4199/256.

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This program will store roster and rating information for a soccer team. coaches rate players during tryouts to ensure a balance
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Answer:

Explanation:

#include <iostream>

#include <vector>

using namespace std;

int main() {

   vector<int> jerseyNumber;

   vector<int> rating;

   int temp;

   for (int i = 1; i <= 5; i++) {

       cout << "Enter player " << i

            << "'s jersey number: ";

       cin >> temp;

       jerseyNumber.push_back(temp);

       cout << "Enter player " << i

            << "'s rating: ";

       cin >> temp;

       rating.push_back(temp);

       cout << endl;

   }

   cout << "ROSTER" << endl;

   for (int i = 0; i < 5; i++)

       cout << "Player " << i + 1 << " -- "

            << "Jersey number: " << jerseyNumber.at(i)

            << ", Rating: " << rating.at(i) << endl;

   char option;

   '

   while (true) {

       cout << "MENU" << endl;

       cout << "a - Add player" << endl;

       cout << "d - Remove player" << endl;

       cout << "u - Update player rating" << endl;

       cout << "r - Output players above a rating"

            << endl;

       cout << "o - Output roster" << endl;

       cout << "q - Quit" << endl << endl;

       cout << "Choose an option: ";

       cin >> option;

       switch (option) {

           case 'a':

           case 'A':

               cout << "Enter a new player's"

                    << "jersey number: ";

               cin >> temp;

               jerseyNumber.push_back(temp);

               cout << "Enter the player's rating: ";

               cin >> temp;

               rating.push_back(temp);

               break;

           case 'd':

           case 'D':

               cout << "Enter a jersey number: ";

               cin >> temp;

               int i;

               for (i = 0; i < jerseyNumber.size();

                    i++) {

                   if (jerseyNumber.at(i) == temp) {

                       jerseyNumber.erase(

                               jerseyNumber.begin() + i);

                       rating.erase(rating.begin() + i);

                       break;

                   }

               }

               break;

           case 'u':

           case 'U':

               cout << "Enter a jersey number: ";

               cin >> temp;

               for (int i = 0; i < jerseyNumber.size();

                    i++) {

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                       cin >> temp;

                       rating.at(i) = temp;

                       break;

                   }

               }

               break;

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               cin >> temp;

               cout << "\nABOVE " << temp << endl;

               for (int i = 0; i < jerseyNumber.size();

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                            << "Jersey number: "

                            << jerseyNumber.at(i)

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                            << rating.at(i) << endl;

               break;

           case 'o':

           case 'O':

               cout << "ROSTER" << endl;

               for (int i = 0; i < jerseyNumber.size();

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                   cout << "Player " << i + 1 << " -- "

                        << "Jersey number: "

                        << jerseyNumber.at(i) << ", Rating: "

                        << rating.at(i) << endl;

               break;

           case 'q':

               return 0;

           default:

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       }

   }

}

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Answer:

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Cash Break-Even

Case 1 = $11,766.67    Case 2 = $1342.86     Case 3 = $168.75

Explanation:

According to the scenario, computation of the given data are as follow:-

Accounting Break-Even = (Fixed Cost + Depreciation Cost) ÷ (Price Unit -Variable Unit)

Case 1 - ($7,060,000 + $1,550,000) ÷ ($3075 - $2,475)

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Case 2 - ( $47,000 + $250,000) ÷ ($96 - $61)

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Cash Break Even = Fixed Cost ÷ (Price Unit - Variable Unit)

Case 1 - $7,060,000 ÷ ($3075 - $2,475)

= $7,060,000 ÷ $600

= $11,766.67

Case 2 - $47,000 ÷ ($96 - $61)

= $47,000 ÷ $35 = $1342.86

Case 3 - $2,700 ÷ ($21 - $5)

= $2,700 ÷ $16 = $168.75

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