To determine which statements are true and which are false, we'll convert the fractions reported for each bird into a common denominator so we can compare them directly.
The findings:
- Blue jay: \( \frac{1}{8} \)
- Finch: \( \frac{6}{7} \)
- Blackbird: \( \frac{1}{3} \)
- Owl: \( \frac{11}{24} \)
The least common multiple (LCM) of the denominators (8, 7, 3, and 24) is 168. Now, we'll convert each fraction to have a denominator of 168.
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Blue jay: \[ \frac{1}{8} = \frac{1 \times 21}{8 \times 21} = \frac{21}{168} \]
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Finch: \[ \frac{6}{7} = \frac{6 \times 24}{7 \times 24} = \frac{144}{168} \]
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Blackbird: \[ \frac{1}{3} = \frac{1 \times 56}{3 \times 56} = \frac{56}{168} \]
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Owl: \[ \frac{11}{24} = \frac{11 \times 7}{24 \times 7} = \frac{77}{168} \]
Now we have the following counts for each type of bird, expressed as follows:
- Blue jay: \( 21 \)
- Finch: \( 144 \)
- Blackbird: \( 56 \)
- Owl: \( 77 \)
Now we can evaluate the statements:
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Over half the students found a finch.
- True: \( 144 > \frac{1}{2} \times 168 = 84 \)
-
More students found a blackbird than a finch.
- False: \( 56 < 144 \)
-
More students found a blue jay than an owl.
- False: \( 21 < 77 \)
-
More students found a finch than a blackbird.
- True: \( 144 > 56 \)
-
More students found a blackbird than an owl.
- False: \( 56 < 77 \)
In summary, the truth values of the statements are:
- Over half the students found a finch: True
- More students found a blackbird than a finch: False
- More students found a blue jay than an owl: False
- More students found a finch than a blackbird: True
- More students found a blackbird than an owl: False