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Brilliant_brown [7]
3 years ago
5

East High School scored a total of 71 points in their final basketball game of the season. The team successfully made a total of

32 baskets. The score was composed of two-point baskets and three-point baskets. The following system of equations can be used to model the number of two-point baskets, x, and the number of three-point baskets, y, that were made during the game. First, determine the point on the graph that represents the solution to the system of equations. Notice that one of the equations in the system has already been graphed. Then, determine the number of two-point baskets and three-point baskets that are made during the final basketball game of the season.
Mathematics
1 answer:
ale4655 [162]3 years ago
0 0

Answer:

Step-by-step explanation:

x = 2 pointers and y = 3 pointers

x + y = 32......x = 32 - y

2x + 3y = 71

2(32 - y) + 3y = 71

64 - 2y + 3y = 71

y = 71 - 64

y = 7

x + y = 32

x + 7 = 32

x = 32 - 7

x = 25

check....

2x + 3y = 71

2(25) + 3(7) = 71

50 + 21 = 71

71 = 71 (correct)..it checks out

so the solution to this system is : (25,7)

there were 25 two points shots made and 7 three point shots made

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Answer:

Objective Minimize 10x1 +15x2 + 12x3+18x4+15x5+ 21x6

The central total cost $ 17586 due to the number of central undergraduate students 1458 is very high.

The total minimum cost would $ 17586 +$260+ $300= $ 18146

Step-by-step explanation:

Let

X1 = # of undergraduate students from the East region,

X2 = # of graduate students from the East region,

X3 = # of undergraduate students from the Central region,

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Then  the cost functions are

y1= 10x1 +15x2

y2= 12x3+18x4

y3= 15x5+ 21x6

According to the given conditions

The constraints are

x1 +x2 + x3+x4+ x5+ x6 ≥ 1500------- A

15X2 +18X4+21X6 ≥ 400---------B

21X6 ≥ 100

X6 ≥ 100/21

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15X2 +18X4+21X6 ≥ 400

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Taking X2= 14

Now putting the values in the cost equations to check whether the conditions are satisfied.

y1= 10x1 +15x2

y1= 10 (5) + 15(14)= 50 + 210= $ 260

y3= 15x5+ 21x6

y3= 15 (13) + 21(5)

y3= 195+105= $ 300

x1 +x2 + x3+x4+ x5+ x6 ≥ 1500

5+14+x3+5+13+5≥ 1500

x3≥ 1500-42

x3≥ 1458

y2= 12x3+18x4

y2= 12 (1458)  + 18 (5)

y2= 17496 +90

y2= $ 17586

The cost can be minimized if the number of students from

                                 Undergraduate         Graduate

East Region              X1≤ 5                            X2  ≥ 13.67

Central                      X3≥ 1458                   X4 ≥ 4.167

West                          X5 ≥ 13                          X6 ≥ 4.76

This will result in the required number of students that is 1500

Constraints:

East Undergraduate must not be greater than 5

East Graduate must not be less than 13

Central Undergraduate must  be greater than 1458

Central Graduate must  be greater than 4

West Undergraduate must  be greater than 13

West Graduate must  be greater than 4

The central total cost $ 17586 due to the number of central undergraduate students 1458 is very high.

The east region has a least cost of $260 and west region has a cost of $300.

The total minimum cost would $ 17586 +$260+ $300= $ 18146

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