Associative operations? Select ALL.
- A. Addition
- B. Subtraction
- C. Multiplication
- D. Division
- E. Exponentiation
Correct Answer & Rationale
Correct Answer: A,C
Associative operations allow the grouping of numbers in different ways without changing the result. Addition (A) and multiplication (C) are associative; for example, (a + b) + c = a + (b + c) and (a × b) × c = a × (b × c). Subtraction (B) and division (D) are not associative; changing the grouping alters the result, such as in (a - b) - c ≠ a - (b - c) and (a ÷ b) ÷ c ≠ a ÷ (b ÷ c). Exponentiation (E) is also not associative, as (a^b)^c ≠ a^(b^c). Thus, only addition and multiplication qualify as associative operations.
Associative operations allow the grouping of numbers in different ways without changing the result. Addition (A) and multiplication (C) are associative; for example, (a + b) + c = a + (b + c) and (a × b) × c = a × (b × c). Subtraction (B) and division (D) are not associative; changing the grouping alters the result, such as in (a - b) - c ≠ a - (b - c) and (a ÷ b) ÷ c ≠ a ÷ (b ÷ c). Exponentiation (E) is also not associative, as (a^b)^c ≠ a^(b^c). Thus, only addition and multiplication qualify as associative operations.
Other Related Questions
Greatest?
- A. 245 thousandths
- B. 24 hundredths
- C. 3 tenths
- D. 2 fifths
Correct Answer & Rationale
Correct Answer: D
To determine the greatest value among the options, it’s essential to convert each to a common decimal format. A: 245 thousandths equals 0.245. B: 24 hundredths equals 0.24. C: 3 tenths equals 0.3. D: 2 fifths equals 0.4 (since 2 divided by 5 is 0.4). Comparing these values, 0.4 (D) is greater than 0.3 (C), 0.24 (B), and 0.245 (A). Thus, option D represents the largest value. Options A, B, and C are all less than D, making them incorrect choices.
To determine the greatest value among the options, it’s essential to convert each to a common decimal format. A: 245 thousandths equals 0.245. B: 24 hundredths equals 0.24. C: 3 tenths equals 0.3. D: 2 fifths equals 0.4 (since 2 divided by 5 is 0.4). Comparing these values, 0.4 (D) is greater than 0.3 (C), 0.24 (B), and 0.245 (A). Thus, option D represents the largest value. Options A, B, and C are all less than D, making them incorrect choices.
Which student wrote the estimate closest to 1,592 + 8?
- A. Isabella
- B. Jayden
- C. Michael
- D. Sarah
Correct Answer & Rationale
Correct Answer: A
Isabella's estimate of 1,592 + 8 is 1,600, which is closest to the actual sum. This estimation rounds 1,592 to 1,590 and adds 10 for simplicity, yielding 1,600. Jayden likely underestimated or rounded incorrectly, resulting in a less accurate estimate. Michael may have rounded too far or added an incorrect value, leading to a larger discrepancy. Sarah's estimate might not have accounted properly for the addition, causing it to stray further from the actual result. Thus, Isabella’s approach demonstrates the most accurate estimation strategy.
Isabella's estimate of 1,592 + 8 is 1,600, which is closest to the actual sum. This estimation rounds 1,592 to 1,590 and adds 10 for simplicity, yielding 1,600. Jayden likely underestimated or rounded incorrectly, resulting in a less accurate estimate. Michael may have rounded too far or added an incorrect value, leading to a larger discrepancy. Sarah's estimate might not have accounted properly for the addition, causing it to stray further from the actual result. Thus, Isabella’s approach demonstrates the most accurate estimation strategy.
Caterpillar 1 ft in 7.5 min. 18 min?
- A. 2.4
- B. 8
- C. 11.5
- D. 25.5
Correct Answer & Rationale
Correct Answer: A
To determine how far the caterpillar travels in 18 minutes, first calculate its speed. It moves 1 foot in 7.5 minutes, which equates to \( \frac{1 \text{ ft}}{7.5 \text{ min}} \). In 18 minutes, the distance covered can be calculated using the formula: \[ \text{Distance} = \text{Speed} \times \text{Time} \] Converting 18 minutes into feet: \[ \text{Distance} = \left(\frac{1 \text{ ft}}{7.5 \text{ min}}\right) \times 18 \text{ min} = 2.4 \text{ ft} \] Option B (8) overestimates the distance, while C (11.5) and D (25.5) significantly exceed the calculated distance, demonstrating a misunderstanding of the speed-time relationship.
To determine how far the caterpillar travels in 18 minutes, first calculate its speed. It moves 1 foot in 7.5 minutes, which equates to \( \frac{1 \text{ ft}}{7.5 \text{ min}} \). In 18 minutes, the distance covered can be calculated using the formula: \[ \text{Distance} = \text{Speed} \times \text{Time} \] Converting 18 minutes into feet: \[ \text{Distance} = \left(\frac{1 \text{ ft}}{7.5 \text{ min}}\right) \times 18 \text{ min} = 2.4 \text{ ft} \] Option B (8) overestimates the distance, while C (11.5) and D (25.5) significantly exceed the calculated distance, demonstrating a misunderstanding of the speed-time relationship.
Answerable?
- A. 4.5 pounds?
- B. At least 15?
- C. Less than 8?
- D. 6-12 pounds?
Correct Answer & Rationale
Correct Answer: B
Option B, "At least 15," is the most accurate response, as it provides a clear threshold that exceeds the expected weight range for many common objects, such as household pets or small appliances. Option A, "4.5 pounds," is too low for many items, making it an unreliable estimate. Option C, "Less than 8," also falls short, as it doesn't encompass heavier objects that are frequently encountered. Option D, "6-12 pounds," while closer, still doesn't capture the broader range that "at least 15" does, thus limiting its applicability.
Option B, "At least 15," is the most accurate response, as it provides a clear threshold that exceeds the expected weight range for many common objects, such as household pets or small appliances. Option A, "4.5 pounds," is too low for many items, making it an unreliable estimate. Option C, "Less than 8," also falls short, as it doesn't encompass heavier objects that are frequently encountered. Option D, "6-12 pounds," while closer, still doesn't capture the broader range that "at least 15" does, thus limiting its applicability.