To determine the truth of the statements based on the proportions given, let's first convert each reported fraction into decimals or percentages for better comparison.
Let \( N \) be the total number of students.
- Blue Jay: \( \frac{1}{8}N = 0.125N \)
- Finch: \( \frac{6}{7}N \approx 0.857N \)
- Blackbird: \( \frac{1}{3}N \approx 0.333N \)
- Owl: \( \frac{11}{24}N \approx 0.458N \)
Now let's analyze each statement:
-
Over half the students found a finch.
- We know \( \frac{6}{7}N \approx 0.857N \) which is greater than half of \( N \) (which is \( 0.5N \)).
- True
-
More students found a blackbird than a finch.
- The number of students who found a blackbird is \( \frac{1}{3}N \approx 0.333N \) and for a finch, it's \( \frac{6}{7}N \approx 0.857N \).
- Since \( 0.333N < 0.857N \),
- False
-
More students found a blue jay than an owl.
- The number of students who found a blue jay is \( \frac{1}{8}N = 0.125N \) and for an owl, it's \( \frac{11}{24}N \approx 0.458N \).
- Since \( 0.125N < 0.458N \),
- False
-
More students found a finch than a blackbird.
- As calculated earlier, \( \frac{6}{7}N \approx 0.857N \) (finch) is greater than \( \frac{1}{3}N \approx 0.333N \) (blackbird).
- True
-
More students found a blackbird than an owl.
- The number of students who found a blackbird is \( \frac{1}{3}N \approx 0.333N \) compared to the number who found an owl \( \frac{11}{24}N \approx 0.458N \).
- Since \( 0.333N < 0.458N \),
- False
Summary of the Statements
- 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