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dlinn [17]
3 years ago
14

A singing contest eliminates contestants after each round. The number of contestants that go into round n is equal to the number

of contestants in the previous round raised to the n-1/n ​-th ​power, rounded to the nearest integer. The number of contestants in Round 2 is 243. How many contestants will be in Round​ 5?
Mathematics
1 answer:
dimaraw [331]3 years ago
8 0

Answer:

9

Step-by-step explanation:

Since we know the number of contestants in Round 2, we could go step by step and calculate the number of remaining contestants in Round 3, then Round 4 and finally Round 5.

However, we can do it without calculating. So in Round 3, there will be:

R3 = R2^((n-1)/n), where n is number of the round

R3 = R2^(2/3)

For Round 4, the number will be:

R4 = (R2^(2/3))^(3/4)

And Round 5 will be:

R5 = ((R2^(2/3))^(3/4))^(4/5)

Threes and fives in exponents will cancel out and we'll remain only with:

R5 = R2^(2/5)

R5 = 249^(2/5)

R5 = 9.08

Rounded to the nearest integer, the number of contestants in Round 5 is 9.

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I can't make a table here but I will try my best to seperate the numbers in groups

4.5 -34.5   :   8,17,23,27,10,23,23,27,5,27

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64.5 -94.5   :   84,88,65,65,65,84

Step-by-step explanation:

In this question there are three categories of intervals and a set of numbers. Essentially, the question wants you to seperate the numbers within each category of interval. I could not make the table here, but it is basically laid out for you. As for the histogram, the question in and out of itself is very vague because there are many ways you can set up a histogram. The file below may not be what you're looking for but it is how I imagined it would be.

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