A bag of dog food weighs 40 pounds. The amount of food in the bag is more than 3 times the amount needed to feed a dog for one week. Which inequality can be used to determine the possible values for p, the pounds of food needed to feed the dog for one week?
- A. p < 3(40)
- B. 3p < 40
- C. p > 3(40)
- D. 3p > 40
Correct Answer & Rationale
Correct Answer: D
To find the amount of food needed for one week, we know that the total weight of the dog food (40 pounds) is more than three times the weekly requirement (3p). Therefore, the relationship can be expressed as 3p < 40, indicating that the total food exceeds three times the weekly amount. Option A (p < 3(40)) incorrectly suggests that the weekly requirement is less than three times the total weight, which is not supported by the problem statement. Option B (3p < 40) misrepresents the relationship, as it implies the total food is less than three times the weekly need, contradicting the given information. Option C (p > 3(40)) inaccurately states that the weekly requirement exceeds three times the total weight, which is impossible given the context. Thus, the correct inequality is 3p > 40, indicating the total food is indeed more than three times the weekly requirement.
To find the amount of food needed for one week, we know that the total weight of the dog food (40 pounds) is more than three times the weekly requirement (3p). Therefore, the relationship can be expressed as 3p < 40, indicating that the total food exceeds three times the weekly amount. Option A (p < 3(40)) incorrectly suggests that the weekly requirement is less than three times the total weight, which is not supported by the problem statement. Option B (3p < 40) misrepresents the relationship, as it implies the total food is less than three times the weekly need, contradicting the given information. Option C (p > 3(40)) inaccurately states that the weekly requirement exceeds three times the total weight, which is impossible given the context. Thus, the correct inequality is 3p > 40, indicating the total food is indeed more than three times the weekly requirement.
Other Related Questions
Dr. Evers is experimenting with light beams and prisms. He passes a beam of white light through a triangular prism which spreads the light out into its six rainbow colors. The bases of the prism are equilateral triangles. The surface area of this prism is 4,292 square millimeters. The area of each triangular face is 271 square millimeters. Which expression can be used to find h, the height, in millimeters, of the prism?
- A. 4,292/3(25)
- B. 4,292/271
- C. (4,292-271)/25
- D. (4,292-2(271))/3(25)
Correct Answer & Rationale
Correct Answer: D
To find the height \( h \) of the prism, we start with the total surface area of the prism, which includes the two triangular bases and three rectangular sides. The area of the two triangular bases is \( 2 \times 271 = 542 \) square millimeters. Subtracting this from the total surface area gives \( 4,292 - 542 = 3,750 \) square millimeters for the area of the rectangular sides. Since the height \( h \) is involved in the area of the rectangles, dividing this area by the perimeter of the base (which is \( 3 \times 25 = 75 \) mm) leads to \( h = \frac{3,750}{75} \) or \( \frac{4,292 - 542}{75} \), simplifying to option D. Options A and B incorrectly compute the height without accounting for the rectangular areas properly. Option C miscalculates the area of the triangular bases and does not consider the full surface area needed to find \( h \). Thus, only option D correctly utilizes the total surface area and the dimensions of the prism to derive the height.
To find the height \( h \) of the prism, we start with the total surface area of the prism, which includes the two triangular bases and three rectangular sides. The area of the two triangular bases is \( 2 \times 271 = 542 \) square millimeters. Subtracting this from the total surface area gives \( 4,292 - 542 = 3,750 \) square millimeters for the area of the rectangular sides. Since the height \( h \) is involved in the area of the rectangles, dividing this area by the perimeter of the base (which is \( 3 \times 25 = 75 \) mm) leads to \( h = \frac{3,750}{75} \) or \( \frac{4,292 - 542}{75} \), simplifying to option D. Options A and B incorrectly compute the height without accounting for the rectangular areas properly. Option C miscalculates the area of the triangular bases and does not consider the full surface area needed to find \( h \). Thus, only option D correctly utilizes the total surface area and the dimensions of the prism to derive the height.
The top speed of the aircraft carrier USS Enterprise is 33 knots. A knot is the speed of a ship in nautical miles per hour. What is the top speed, in miles per hour? (1 nautical mile = 6,076 feet; 1 mile - 5,280 feet)
- A. 24 miles per hour
- B. 38 miles per hour
- C. 33 miles per hour
- D. 29 miles per hour
Correct Answer & Rationale
Correct Answer: B
To convert knots to miles per hour, it’s essential to understand the relationship between nautical miles and standard miles. Since 1 nautical mile equals 6,076 feet and 1 mile equals 5,280 feet, we can set up the conversion: 1 nautical mile = 6,076 feet / 5,280 feet/mile = 1.151 miles. Thus, to convert 33 knots to miles per hour: 33 knots × 1.151 miles/nautical mile = 38.0 miles per hour. Option A (24 mph) is too low, failing to account for the conversion factor. Option C (33 mph) incorrectly assumes knots and miles per hour are equivalent. Option D (29 mph) underestimates the conversion, not reaching the correct calculation.
To convert knots to miles per hour, it’s essential to understand the relationship between nautical miles and standard miles. Since 1 nautical mile equals 6,076 feet and 1 mile equals 5,280 feet, we can set up the conversion: 1 nautical mile = 6,076 feet / 5,280 feet/mile = 1.151 miles. Thus, to convert 33 knots to miles per hour: 33 knots × 1.151 miles/nautical mile = 38.0 miles per hour. Option A (24 mph) is too low, failing to account for the conversion factor. Option C (33 mph) incorrectly assumes knots and miles per hour are equivalent. Option D (29 mph) underestimates the conversion, not reaching the correct calculation.
The daily cost, C(x), tor a company to produce x microscopes is given by the equation C(x) = 300 + 10.5x. What is the cost of producing 50 microscopes?
- A. $41,250
- B. $360.50
- C. $15,525
- D. $825
Correct Answer & Rationale
Correct Answer: D
To find the cost of producing 50 microscopes, substitute x = 50 into the cost equation C(x) = 300 + 10.5x. This yields C(50) = 300 + 10.5(50), resulting in C(50) = 300 + 525 = 825. Thus, the cost for 50 microscopes is $825. Option A ($41,250) is incorrect as it likely results from a miscalculation or misunderstanding of the equation. Option B ($360.50) underestimates the production cost by omitting the correct multiplication factor. Option C ($15,525) suggests an error in the calculation, possibly misinterpreting the coefficients in the equation.
To find the cost of producing 50 microscopes, substitute x = 50 into the cost equation C(x) = 300 + 10.5x. This yields C(50) = 300 + 10.5(50), resulting in C(50) = 300 + 525 = 825. Thus, the cost for 50 microscopes is $825. Option A ($41,250) is incorrect as it likely results from a miscalculation or misunderstanding of the equation. Option B ($360.50) underestimates the production cost by omitting the correct multiplication factor. Option C ($15,525) suggests an error in the calculation, possibly misinterpreting the coefficients in the equation.
The radius of the sphere below is 6 centimeters (cm). What is the volume, in cubic centimeters, of the sphere?
- A. 904.32
- B. 150.72
- C. 25.12
- D. 75.36
Correct Answer & Rationale
Correct Answer: A
To find the volume of a sphere, the formula \( V = \frac{4}{3} \pi r^3 \) is used, where \( r \) is the radius. For a radius of 6 cm, the calculation is: \[ V = \frac{4}{3} \pi (6)^3 = \frac{4}{3} \pi (216) \approx 904.32 \, \text{cm}^3 \] Option A (904.32) correctly represents this volume. Option B (150.72) and Option C (25.12) are significantly lower than the actual volume, indicating miscalculations or incorrect application of the formula. Option D (75.36) is also incorrect, as it does not appropriately reflect the cubic growth of the volume with respect to the radius, resulting in an underestimation.
To find the volume of a sphere, the formula \( V = \frac{4}{3} \pi r^3 \) is used, where \( r \) is the radius. For a radius of 6 cm, the calculation is: \[ V = \frac{4}{3} \pi (6)^3 = \frac{4}{3} \pi (216) \approx 904.32 \, \text{cm}^3 \] Option A (904.32) correctly represents this volume. Option B (150.72) and Option C (25.12) are significantly lower than the actual volume, indicating miscalculations or incorrect application of the formula. Option D (75.36) is also incorrect, as it does not appropriately reflect the cubic growth of the volume with respect to the radius, resulting in an underestimation.