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inysia [295]
3 years ago
14

The graph represents Jason's speed on a trip. What is the rate of change for the period of time between 10 and 20 minutes?

Mathematics
2 answers:
11111nata11111 [884]3 years ago
7 0
From 10 minutes to 20 minutes there is no rate of change, it stays at a constant and steady rate

Your answer would be A) 0
natali 33 [55]3 years ago
6 0

Answer:

Option A - 0

Step-by-step explanation:

Given : The graph represents Jason's speed on a trip.

To find : What is the rate of change for the period of time between 10 and 20 minutes?                        

Solution :

The rate of change is defined as the division of change in  speed by change in time.

Now, the rate of change for the period of time between 10 and 20 minutes is

Speed in 10 minutes  is 50 mph

Speed in 20 minutes is 50 mph

The rate of change is r=\frac{s_2-s_1}{t_2-t_1}

r=\frac{50-50}{20-10}

r=0

Therefore, Option A is correct.

The rate of change for the period of time between 10 and 20 minute is 0 or constant as shown graphically.

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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,

X4 = # of graduate students from the Central region,

X5 = # of undergraduate students from the West region, and

X6 = # of graduate students from the West region.

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

X6 ≥ 4.76

Taking

X6= 5

10X1 ≤ 500

X1 ≤ 500/10

X1≤ 5

18X4 ≥ 75

X4 ≥ 75/18

X4 ≥ 4.167

Taking

X4= 5

Putting the values

15X5+ 21X6 ≥ 300

15X5+ 21(5) ≥ 300

15X5+ 105 ≥ 300

15X5 ≥ 300-105

15X5 ≥ 195

X5 ≥ 195/15

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Putting value of X6 and X4 in B

15X2 +18X4+21X6 ≥ 400

15X2 +18(5)+21(5) ≥ 400

15X2 +195 ≥ 400

15X2  ≥ 400-195

15X2  ≥ 205

X2  ≥ 205/15

X2  ≥ 13.67

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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