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nikitadnepr [17]
3 years ago
12

If the product of two positive fractions a and b is 15/56, find three pairs of possible values for a and b

Mathematics
1 answer:
Andrews [41]3 years ago
4 0

Answer:

First possible pairs = (3/7)*(5/8) = 15/56

second possible pairs = (3/4)*(5/14) = 15/56

third possible pairs = (1/2)*(15/28) = 15/56

Step-by-step explanation:

According to the question,

a x b = 15/56

three possible pairs will be as follows:

Before proceeding to make possible pairs, we have to know the prime factorization of the numbers -

15 = 1, 3, 5, 15

56 = 1, 2, 4, 7, 8, 14, 28, 56

As we need three possible pairs, we have to check which pairs take us to 15/56. Again, since the fraction is positive, therefore, the fractions will be proper. Therefore, 3/2, or 5/4 will not be counted. Therefore,

First possible pairs = (3/7)*(5/8) = 15/56

second possible pairs = (3/4)*(5/14) = 15/56

third possible pairs = (1/2)*(15/28) = 15/56

fourth possible pairs = (3/8)*(5/7) = 15/56

So, we can get those four possible pairs. Among those, first 3 pairs are different. Therefore, those are possible pairs.

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Let us check series : -10 /-2 = 5 -50/-10 = 5

so it is a geometric series with a= -2 and r= 5

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A survey of 1,107 tourists visiting Orlando was taken. Of those surveyed:
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Answer:

602 tourists visited only the LEGOLAND.

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 = 58

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 1,107 tourists suveyed. This means that:

a + b + c + d + (A \cap B) + (A \cap C) + (B \cap C) + (A \cap B \cap C) = 1,107

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

The problem states that:

36 tourists had visited all three theme parks. So:

(A \cap B \cap C) = 36

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

(A \cap B) + (A \cap B \cap C) = 72

(A \cap B) = 72 - 36

(A \cap B) = 36

79 tourists had visited both the Magic Kingdom and Universal Studios

(B \cap C) + (A \cap B \cap C) = 79

(B \cap C) = 79 - 36

(B \cap C) = 43

68 tourists had visited both the Magic Kingdom and LEGOLAND

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

(A \cap C) = 68 - 36

(A \cap C) = 32

258 tourists had visited Universal Studios:

B = 258

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

258 = b + 43 + 36 + 36

b = 143

268 tourists had visited the Magic Kingdom:

C = 268

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

268 = c + 32 + 43 + 36

c = 157

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

We have to find the value of a, and we can do this by the following equation:

a + b + c + d + (A \cap B) + (A \cap C) + (B \cap C) + (A \cap B \cap C) = 1,107

a + 143 + 157 + 58 + 36 + 32 + 43 + 36 = 1,107

a = 602

602 tourists visited only the LEGOLAND.

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