To determine the truthfulness of each statement, we first need to convert the fractions of students who found each type of bird into a common denominator. The denominators are 8, 7, 3, and 24. The least common multiple (LCM) of 8, 7, 3, and 24 is 168.
Next, we calculate the number of students who found each bird type by converting each fraction to an equivalent fraction with a denominator of 168.
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Blue Jay: \( \frac{1}{8} \) of the students. \[ \frac{1}{8} \times \frac{21}{21} = \frac{21}{168} \]
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Finch: \( \frac{6}{7} \) of the students. \[ \frac{6}{7} \times \frac{24}{24} = \frac{144}{168} \]
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Blackbird: \( \frac{1}{3} \) of the students. \[ \frac{1}{3} \times \frac{56}{56} = \frac{56}{168} \]
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Owl: \( \frac{11}{24} \) of the students. \[ \frac{11}{24} \times \frac{7}{7} = \frac{77}{168} \]
Now we have the following counts of students for each bird type based on a total of 168 students:
- Blue Jay: 21 students
- Finch: 144 students
- Blackbird: 56 students
- Owl: 77 students
Now, let's evaluate each statement:
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Over half the students found a finch. (True)
- \( \frac{144}{168} \) is greater than \( \frac{84}{168} \) (which is half of 168).
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More students found a blackbird than a finch. (False)
- 56 (blackbirds) is less than 144 (finches).
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More students found a blue jay than an owl. (False)
- 21 (blue jays) is less than 77 (owls).
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More students found a finch than a blackbird. (True)
- 144 (finches) is greater than 56 (blackbirds).
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More students found a blackbird than an owl. (False)
- 56 (blackbirds) is less than 77 (owls).
Final results:
- 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)
11 answers