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Answer:
170 children
74 students
85 adults
Step-by-step explanation:
Given
Let:

For the capacity, we have:

For the tickets sold, we have:

Half as many as adults are children implies that:

Required
Solve for A, C and S
The equations to solve are:
-- (1)
-- (2)
-- (3)
Make C the subject in (3)

Substitute
in (1) and (2)
-- (1)


Make S the subject

-- (2)



Substitute 



Solve for A


Recall that: 


Recall that: 



Hence, the result is:



Answer:
- <u>It is a combination</u>
<u></u>
Explanation:
This is a combination because the order does not matter: the Beagle - Collie couple is the same as the Collie - Beagle couple, so it is only counted once.
The formula for combinations is:

For five diffferent dogs to be put in pairs, the number of different pairs is:

Area of cone = 1/3 base × height = 1/3 × 8 × 10 = approximately 26.7 m
Ans= approx. 26.4 m
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