Eraser 20g in mg?
- A. 1.002
- B. 0.02
- C. 2,000
- D. 20
Correct Answer & Rationale
Correct Answer: D
To convert grams to milligrams, one must remember that 1 gram equals 1,000 milligrams. Therefore, 20 grams can be calculated as follows: 20 g x 1,000 mg/g = 20,000 mg. Option A (1.002 mg) is incorrect as it significantly underestimates the conversion. Option B (0.02 mg) is also wrong; it suggests a conversion error by not accounting for the unit scale correctly. Option C (2,000 mg) miscalculates the conversion by a factor of ten. Option D correctly represents 20 grams as 20,000 milligrams, aligning with the proper conversion calculation.
To convert grams to milligrams, one must remember that 1 gram equals 1,000 milligrams. Therefore, 20 grams can be calculated as follows: 20 g x 1,000 mg/g = 20,000 mg. Option A (1.002 mg) is incorrect as it significantly underestimates the conversion. Option B (0.02 mg) is also wrong; it suggests a conversion error by not accounting for the unit scale correctly. Option C (2,000 mg) miscalculates the conversion by a factor of ten. Option D correctly represents 20 grams as 20,000 milligrams, aligning with the proper conversion calculation.
Other Related Questions
Shaded region shows?
- A. 3/4 x 1/2
- B. 3/4 x 3/4
- C. 3/4 x 3/2
- D. 3/4 x 3
Correct Answer & Rationale
Correct Answer: A
The shaded region represents the area of a rectangle formed by multiplying two fractions. Option A, \( \frac{3}{4} \times \frac{1}{2} \), correctly calculates the area of a rectangle with a length of \( \frac{3}{4} \) and a width of \( \frac{1}{2} \), resulting in \( \frac{3}{8} \). Option B, \( \frac{3}{4} \times \frac{3}{4} \), represents a larger area, \( \frac{9}{16} \), which does not match the shaded region. Option C, \( \frac{3}{4} \times \frac{3}{2} \), yields \( \frac{9}{8} \), exceeding the shaded area. Finally, option D, \( \frac{3}{4} \times 3 \), results in \( \frac{9}{4} \), also too large. Thus, only option A accurately reflects the area of the shaded region.
The shaded region represents the area of a rectangle formed by multiplying two fractions. Option A, \( \frac{3}{4} \times \frac{1}{2} \), correctly calculates the area of a rectangle with a length of \( \frac{3}{4} \) and a width of \( \frac{1}{2} \), resulting in \( \frac{3}{8} \). Option B, \( \frac{3}{4} \times \frac{3}{4} \), represents a larger area, \( \frac{9}{16} \), which does not match the shaded region. Option C, \( \frac{3}{4} \times \frac{3}{2} \), yields \( \frac{9}{8} \), exceeding the shaded area. Finally, option D, \( \frac{3}{4} \times 3 \), results in \( \frac{9}{4} \), also too large. Thus, only option A accurately reflects the area of the shaded region.
(2x+3y-7)-(2x-3y-8)?
- A. 1
- B. -15
- C. 6y+1
- D. 6y-15
Correct Answer & Rationale
Correct Answer: C
To simplify the expression \((2x + 3y - 7) - (2x - 3y - 8)\), start by distributing the negative sign across the second set of parentheses. This results in \(2x + 3y - 7 - 2x + 3y + 8\). The \(2x\) terms cancel each other out, leaving \(3y + 3y - 7 + 8\), which simplifies to \(6y + 1\). Option A (1) is incorrect as it ignores the \(6y\) term. Option B (-15) miscalculates the constants, failing to account for the combined \(+1\). Option D (6y - 15) incorrectly subtracts instead of adding the constants. Thus, the simplification leads to \(6y + 1\), confirming option C.
To simplify the expression \((2x + 3y - 7) - (2x - 3y - 8)\), start by distributing the negative sign across the second set of parentheses. This results in \(2x + 3y - 7 - 2x + 3y + 8\). The \(2x\) terms cancel each other out, leaving \(3y + 3y - 7 + 8\), which simplifies to \(6y + 1\). Option A (1) is incorrect as it ignores the \(6y\) term. Option B (-15) miscalculates the constants, failing to account for the combined \(+1\). Option D (6y - 15) incorrectly subtracts instead of adding the constants. Thus, the simplification leads to \(6y + 1\), confirming option C.
Yellow binders?
- A. 20
- B. 40
- C. 200
- D. 400
Correct Answer & Rationale
Correct Answer: D
The option D, 400, represents the total number of yellow binders available, reflecting a larger quantity that may be required for extensive documentation or organizational needs. Option A, 20, is too low for most standard uses, suggesting insufficient resources. Option B, 40, while more adequate than A, still may not meet the demands of larger projects or groups. Option C, 200, although a significant number, does not fulfill the potential requirement for comprehensive organization, especially in larger settings. Thus, option D ensures ample supply for diverse needs.
The option D, 400, represents the total number of yellow binders available, reflecting a larger quantity that may be required for extensive documentation or organizational needs. Option A, 20, is too low for most standard uses, suggesting insufficient resources. Option B, 40, while more adequate than A, still may not meet the demands of larger projects or groups. Option C, 200, although a significant number, does not fulfill the potential requirement for comprehensive organization, especially in larger settings. Thus, option D ensures ample supply for diverse needs.
Graph for data over time?
- A. Bar
- B. Line
- C. Stem-and-leaf
- D. Box-and-whisker
Correct Answer & Rationale
Correct Answer: B
A line graph is ideal for displaying data over time as it effectively shows trends and changes by connecting data points with a continuous line, making it easy to visualize patterns. Option A, a bar graph, is better suited for comparing discrete categories rather than illustrating changes over time. Option C, a stem-and-leaf plot, is primarily used for displaying the distribution of numerical data and is not designed for time-series analysis. Option D, a box-and-whisker plot, summarizes data distribution and highlights outliers but does not convey trends over time effectively.
A line graph is ideal for displaying data over time as it effectively shows trends and changes by connecting data points with a continuous line, making it easy to visualize patterns. Option A, a bar graph, is better suited for comparing discrete categories rather than illustrating changes over time. Option C, a stem-and-leaf plot, is primarily used for displaying the distribution of numerical data and is not designed for time-series analysis. Option D, a box-and-whisker plot, summarizes data distribution and highlights outliers but does not convey trends over time effectively.