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zimovet [89]
3 years ago
14

A fruit company delivers its fruit in two types of boxes: large and small. A delivery of2 large boxes and 12 small boxes has a t

otal weight of 199 kilograms. A delivery of 5 large boxes and 3 small boxes has a total weight of 133 kilograms. How much does each type of box weigh?
Weight of each large box:____ Kilogram(s)
Weight of each small box: ____ Kilogram(s)
Mathematics
1 answer:
kap26 [50]3 years ago
8 0

Answer:

x=wt of small box     y = large box wt

 

First delivery  

   5x  + 2y = 96      solve for y = (96-5x)/2

 

Second Delivery

  8y + 3x = 163       Substitute for y

  8 ( ( 96-5x)/2 ) + 3x = 163       and solve for x

   384 - 20x +3x = 163

17x=221

x=13   =   small box wt   kg               y= (96-5x)/2 = large box wt = 15.5 kg

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Answer:

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3 years ago
Give an example of a value that could be represented by D
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A survey of 1,168 tourists visiting Orlando was taken. Of those surveyed:
sesenic [268]

Answer:

624 tourists only visited the Magic Kindgom.

Step-by-step explanation:

To solve this problem, we must build the Venn's Diagram of this set.

I am going to say that:

-The set A represents the tourists that visited LEGOLAND

-The set B represents the tourists that visited Universal Studios

-The set C represents the tourists that visited Magic Kingdown.

-The value d is the number of tourists that did not visit any of these parks, so: d = 74

We have that:

A = a + (A \cap B) + (A \cap C) + (A \cap B \cap C)

In which a is the number of tourists that only visited LEGOLAND, A \cap B is the number of tourists that visited both LEGOLAND and Universal Studies, A \cap C is the number of tourists that visited both LEGOLAND and the Magic Kingdom. and A \cap B \cap C is the number of students that visited all these parks.

By the same logic, we have:

B = b + (B \cap C) + (A \cap B) + (A \cap B \cap C)

C = c + (A \cap C) + (B \cap C) + (A \cap B \cap C)

This diagram has the following subsets:

a,b,c,d,(A \cap B), (A \cap C), (B \cap C), (A \cap B \cap C)

There were 1,168 tourists suveyed. This means that:

a + b + c + d + (A \cap B) + (A \cap C) + (B \cap C) + (A \cap B \cap C) = 1,168

We start finding the values from the intersection of three sets.

The problem states that:

16 tourists had visited all three theme parks. So:

A \cap B \cap C = 16

91 tourists had visited both LEGOLAND and Universal Studios. So:

(A \cap B) + (A \cap B \cap C) = 91

(A \cap B) = 91-16

(A \cap B) = 75

68 tourists had visited both the Magic Kingdom and Universal Studios. So

(B \cap C) + (A \cap B \cap C) = 68

(B \cap C) = 68-16

(B \cap C) = 52

87 tourists had visited both the Magic Kingdom and LEGOLAND

(A \cap C) + (A \cap B \cap C) = 87

(A \cap C) = 87-16

(A \cap C) = 71

295 tourists had visited Universal Studios

B = 295

B = b + (B \cap C) + (A \cap B) + (A \cap B \cap C)

295 = b + 52 + 75 + 16

b + 143 = 295

b = 152

266 tourists had visited LEGOLAND

A = 266

A = a + (A \cap B) + (A \cap C) + (A \cap B \cap C)

266 = a + 75 + 71 + 16

a + 162 = 266

a = 104

How many tourists only visited the Magic Kingdom (of these three)?

This is the value of c, the we can find in the following equation:

a + b + c + d + (A \cap B) + (A \cap C) + (B \cap C) + (A \cap B \cap C) = 1,168

104 + 152 + c + 74 + 75 + 71 + 52 + 16 = 1,168

c + 544 = 1,168

c = 624

624 tourists only visited the Magic Kindgom.

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