Imagine you had 40 one-metre sections of fencing.
1 answer:
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Answer:
There are 13 families had a parakeet only
Step-by-step explanation:
* Lets explain the problem
- There are 180 families
- 67 families had a dog
- 52 families had a cat
- 22 families had a dog and a cat
- 70 had neither a cat nor a dog, and in addition did not have a
parakeet
- 4 had a cat, a dog, and a parakeet (4 is a part of 22 and 22 is a part
of 67 and 520
* We will explain the Venn-diagram
- A rectangle represent the total of the families
- Three intersected circles:
C represented the cat
D represented the dog
P represented the parakeet
- The common part of the three circle had 4 families
- The common part between the circle of the cat and the circle of the
dog only had 22 - 4 = 18 families
- The common part between the circle of the dog and the circle of the
parakeet only had a families
- The common part between the circle of the cat and the circle of the
parakeet only had b families
- The non-intersected part of the circle of the dog had 67 - 22 - a =
45 - a families
had dogs only
- The non-intersected part of the circle of the cat had 52 - 22 - b =
30 - b families
had cats only
- The non-intersected part of the circle of the parakeet had c families
had parakeets only
- The part out side the circles and inside the triangle has 70 families
- Look to the attached graph for more under stand
∵ The total of the families is 180
∴ The sum of all steps above is 180
∴ 45 - a + 18 + 4 + 30 - b + b + c + a + 70 = 180 ⇒ simplify
- (-a) will cancel (a) and (-b) will cancel (b)
∴ (45 + 18 + 4 + 30 + 70) + (-a + a) + (-b + b) + c = 180
∴ 167 + c = 180 ⇒ subtract 167 from both sides
∴ c = 180 - 167 = 13 families
* There are 13 families had a parakeet only
