The recommended dosage of a medicine is 4 milligrams per kilogram of body weight. What is the recommended dosage, in milligrams, for a person who weighs 84 kilograms?
- A. 21
- B. 88
- C. 324
- D. 336
- E. 2100
Correct Answer & Rationale
Correct Answer: D
To determine the recommended dosage for a person weighing 84 kilograms, multiply their weight by the dosage per kilogram: 4 mg/kg × 84 kg = 336 mg. Option A (21 mg) is incorrect as it significantly underestimates the dosage based on the weight. Option B (88 mg) also miscalculates the dosage, failing to apply the correct multiplication. Option C (324 mg) is close but still incorrect, as it does not reflect the accurate calculation. Option E (2100 mg) is far too high, indicating a misunderstanding of the dosage per kilogram. Thus, 336 mg is the correct dosage for the individual.
To determine the recommended dosage for a person weighing 84 kilograms, multiply their weight by the dosage per kilogram: 4 mg/kg × 84 kg = 336 mg. Option A (21 mg) is incorrect as it significantly underestimates the dosage based on the weight. Option B (88 mg) also miscalculates the dosage, failing to apply the correct multiplication. Option C (324 mg) is close but still incorrect, as it does not reflect the accurate calculation. Option E (2100 mg) is far too high, indicating a misunderstanding of the dosage per kilogram. Thus, 336 mg is the correct dosage for the individual.
Other Related Questions
Which of the following expressions is equivalent to (4x²)(5x³)?
- A. 9xâµ
- B. 9xâ¶
- C. 20xâµ
- D. 20xâ¶
- E. 20xâ¹
Correct Answer & Rationale
Correct Answer: C
To find the equivalent expression for (4x²)(5x³), multiply the coefficients (4 and 5) and add the exponents of x (2 and 3). Thus, 4 × 5 equals 20, and x² × x³ results in x^(2+3) = x⁵. This gives us 20x⁵. Option A (9x⁶) is incorrect because it miscalculates both the coefficient and the exponent. Option B (9x⁷) also miscalculates both the coefficient and exponent. Option D (20x⁶) correctly identifies the coefficient but incorrectly adds the exponents. Option E (20x¹) miscalculates the exponent entirely. Only option C accurately represents the expression as 20x⁵.
To find the equivalent expression for (4x²)(5x³), multiply the coefficients (4 and 5) and add the exponents of x (2 and 3). Thus, 4 × 5 equals 20, and x² × x³ results in x^(2+3) = x⁵. This gives us 20x⁵. Option A (9x⁶) is incorrect because it miscalculates both the coefficient and the exponent. Option B (9x⁷) also miscalculates both the coefficient and exponent. Option D (20x⁶) correctly identifies the coefficient but incorrectly adds the exponents. Option E (20x¹) miscalculates the exponent entirely. Only option C accurately represents the expression as 20x⁵.
sqrt(45) is between what two consecutive whole numbers?
- A. 4 and 5
- B. 5 and 6
- C. 6 and 7
- D. 14 and 15
- E. 22 and 23
Correct Answer & Rationale
Correct Answer: C
To determine between which two consecutive whole numbers \(\sqrt{45}\) lies, we can evaluate the squares of whole numbers around it. Calculating, \(6^2 = 36\) and \(7^2 = 49\). Since \(36 < 45 < 49\), it follows that \(6 < \sqrt{45} < 7\). Therefore, \(\sqrt{45}\) is between 6 and 7. Option A (4 and 5) is incorrect as \(4^2 = 16\) and \(5^2 = 25\), which are both less than 45. Option B (5 and 6) is also wrong since \(5^2 = 25\) and \(6^2 = 36\) are still below 45. Option D (14 and 15) and Option E (22 and 23) are far too high, as \(14^2 = 196\) and \(22^2 = 484\) exceed 45.
To determine between which two consecutive whole numbers \(\sqrt{45}\) lies, we can evaluate the squares of whole numbers around it. Calculating, \(6^2 = 36\) and \(7^2 = 49\). Since \(36 < 45 < 49\), it follows that \(6 < \sqrt{45} < 7\). Therefore, \(\sqrt{45}\) is between 6 and 7. Option A (4 and 5) is incorrect as \(4^2 = 16\) and \(5^2 = 25\), which are both less than 45. Option B (5 and 6) is also wrong since \(5^2 = 25\) and \(6^2 = 36\) are still below 45. Option D (14 and 15) and Option E (22 and 23) are far too high, as \(14^2 = 196\) and \(22^2 = 484\) exceed 45.
One online movie-streaming service costs $8 per month and charges $1.50 per movie. A second service costs $2 per month and charges $2 per movie. For what number of movies per month is the monthly cost of both services the same?
- A. 3
- B. 6
- C. 5
- D. 12
- E. 20
Correct Answer & Rationale
Correct Answer: D
To determine when the costs of both services are equal, we can set up equations based on the monthly fees and per-movie charges. For the first service: Cost = $8 + $1.50 * number of movies (m) Cost = $8 + 1.5m For the second service: Cost = $2 + $2 * number of movies (m) Cost = $2 + 2m Setting the two equations equal gives us: $8 + 1.5m = $2 + 2m Rearranging leads to: $6 = 0.5m m = 12 Thus, when 12 movies are rented, the costs are equal. Options A (3), B (6), and C (5) yield different costs, as they do not satisfy the equation. Option E (20) results in a higher cost for the second service, confirming that 12 is the only solution where both services cost the same.
To determine when the costs of both services are equal, we can set up equations based on the monthly fees and per-movie charges. For the first service: Cost = $8 + $1.50 * number of movies (m) Cost = $8 + 1.5m For the second service: Cost = $2 + $2 * number of movies (m) Cost = $2 + 2m Setting the two equations equal gives us: $8 + 1.5m = $2 + 2m Rearranging leads to: $6 = 0.5m m = 12 Thus, when 12 movies are rented, the costs are equal. Options A (3), B (6), and C (5) yield different costs, as they do not satisfy the equation. Option E (20) results in a higher cost for the second service, confirming that 12 is the only solution where both services cost the same.
When Henry plays the songs on the playlist in a random order, what is the probability a rock song will be played first?
- A. 3/4
- B. 1/3
- C. 1/4
- D. 3/10
- E. 5/16
Correct Answer & Rationale
Correct Answer: D
To find the probability of a rock song being played first, we need to know the total number of songs and how many of those are rock songs. If there are 3 rock songs and a total of 10 songs, the probability is calculated as the number of favorable outcomes (rock songs) divided by the total outcomes (all songs). Thus, the probability is 3/10, which corresponds to option D. Option A (3/4) overestimates the likelihood by implying a much higher proportion of rock songs. Option B (1/3) incorrectly assumes there are fewer total songs than there actually are. Option C (1/4) underrepresents the rock songs available. Option E (5/16) is irrelevant as it does not align with the total number of songs.
To find the probability of a rock song being played first, we need to know the total number of songs and how many of those are rock songs. If there are 3 rock songs and a total of 10 songs, the probability is calculated as the number of favorable outcomes (rock songs) divided by the total outcomes (all songs). Thus, the probability is 3/10, which corresponds to option D. Option A (3/4) overestimates the likelihood by implying a much higher proportion of rock songs. Option B (1/3) incorrectly assumes there are fewer total songs than there actually are. Option C (1/4) underrepresents the rock songs available. Option E (5/16) is irrelevant as it does not align with the total number of songs.