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saul85 [17]
2 years ago
15

A football coach is trying to decide: When a team is ahead late in the game,

Mathematics
1 answer:
nevsk [136]2 years ago
4 0

The better strategy is to play the 'regular” defense. Because the probability of play in the 'regular” defense is more.

<h3>What is probability?</h3>

Its basic premise is that something will almost certainly happen. The percentage of favorable events to the total number of occurrences.

A football coach is trying to decide:

The coach reviews the outcomes of 100 games.

                        Regular defense       Prevent defense       Total

Win                           41                                32                       73

Loss                           9                                18                        27

Total                         50                               50                      100

The probability of winning when playing regular defense will be

⇒ 41 / 50

⇒ 0.82

The probability of winning when playing prevent defense will be

⇒ 32 / 50

⇒ 0.64

Then the better strategy is to play the 'regular” defense.

More about the probability link is given below.

brainly.com/question/795909

#SPJ1

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3

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Function: Yes or no<br> {(-4,6),(-3,2),(1,0), (7,6), (8,2)}
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2 years ago
The graph shows the feasible region for the system with constraints:
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Answer:

The vertices feasible region are (0 , 15) , (10 , 15) , (20 , 5)

The minimum value of the objective function C is 125

Step-by-step explanation:

* Lets look to the graph to answer the question

- There are 3 inequalities

# y ≤ 15 represented by horizontal line (purple line) and cut the

  y-axis at point (0 , 15)

# x + y ≤ 25 represented by a line (green line) and intersected the

  x-axis at point (25 , 0) and the y- axis at point (0 , 25)

# x + 2y ≥ 30 represented by a line (blue line) and intersected the

  x-axis at point (30 , 0) and the y-axis at point (0 , 15)

- The three lines intersect each other in three points

# The blue and purple lines intersected in point (0 , 15)

# The green and the purple lines intersected in point (10 , 15)

# The green and the blue lines intersected in point (20 , 5)

- The three lines bounded the feasible region

∴ The vertices feasible region are (0 , 15) , (10 , 15) , (20 , 5)

- To find the minimum value of the objective function C = 4x + 9y,

  substitute the three vertices of the feasible region in C and chose

  the least answer

∵ C = 4x + 9y

- Use point (0 , 15)

∴ C = 4(0) + 9(15) = 0 + 135 = 135

- Use point (10 , 15)

∴ C = 4(10) + 9(15) = 40 + 135 = 175

- Use point (20 , 5)

∴ C = 4(40) + 9(5) = 80 + 45 = 125

- From all answers the least value is 125

∴ The minimum value of the objective function C is 125

6 0
3 years ago
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gayaneshka [121]
I could be wrong but I got 28. I devided 420 by 15
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