A diver jumps from a platform. The height, h meters, the diver is above the water t seconds after jumping is represented by h = -16t^2 + 16t + 6.5. To the near hundredth of a second, how many seconds after jumping is the diver 2.5 meters above the water?
- A. 2.79
- B. 1.32
- C. 2.83
- D. 1.21
Correct Answer & Rationale
Correct Answer: D
To find when the diver is 2.5 meters above the water, substitute h = 2.5 into the equation: \[ 2.5 = -16t^2 + 16t + 6.5. \] Rearranging gives: \[ -16t^2 + 16t + 4 = 0. \] Using the quadratic formula, we solve for t, yielding two potential solutions. The option D (1.21 seconds) is valid as it falls within the realistic time frame of the jump. Options A (2.79) and C (2.83) exceed the expected time of descent, while B (1.32) does not satisfy the equation, confirming that only D accurately represents the diver's position at 2.5 meters above the water.
To find when the diver is 2.5 meters above the water, substitute h = 2.5 into the equation: \[ 2.5 = -16t^2 + 16t + 6.5. \] Rearranging gives: \[ -16t^2 + 16t + 4 = 0. \] Using the quadratic formula, we solve for t, yielding two potential solutions. The option D (1.21 seconds) is valid as it falls within the realistic time frame of the jump. Options A (2.79) and C (2.83) exceed the expected time of descent, while B (1.32) does not satisfy the equation, confirming that only D accurately represents the diver's position at 2.5 meters above the water.
Other Related Questions
A manufacturing plant makes dog toys in the shape of a sphere. The diameter of each dog toy is 3 inches. What is the surface area, in square inches of each dog toy?
- A. 113.04
- B. 75.36
- C. 28.26
- D. 37.68
Correct Answer & Rationale
Correct Answer: C
To find the surface area of a sphere, the formula used is \(4\pi r^2\). Given the diameter of the dog toy is 3 inches, the radius \(r\) is half of that, which is 1.5 inches. Plugging this into the formula: \[ Surface Area = 4\pi (1.5)^2 = 4\pi (2.25) \approx 28.26 \text{ square inches.} \] Option A (113.04) results from incorrectly using the diameter instead of the radius. Option B (75.36) arises from miscalculating the radius or misapplying the formula. Option D (37.68) likely results from a miscalculation of the surface area formula, possibly using an incorrect value for \(r\).
To find the surface area of a sphere, the formula used is \(4\pi r^2\). Given the diameter of the dog toy is 3 inches, the radius \(r\) is half of that, which is 1.5 inches. Plugging this into the formula: \[ Surface Area = 4\pi (1.5)^2 = 4\pi (2.25) \approx 28.26 \text{ square inches.} \] Option A (113.04) results from incorrectly using the diameter instead of the radius. Option B (75.36) arises from miscalculating the radius or misapplying the formula. Option D (37.68) likely results from a miscalculation of the surface area formula, possibly using an incorrect value for \(r\).
A scale drawing of a truck has a length of 3 inches (in.), as shown below. The actual truck has a length of 18 feet (ft). What scale was used for the drawing?
- A. 6 in. = 1 ft
- B. 1 in. = 15 ft
- C. 1 in. = 6 ft
- D. 15 in. = 1 ft
Correct Answer & Rationale
Correct Answer: C
To determine the scale used for the drawing, we first convert the actual truck length from feet to inches. Since 1 foot equals 12 inches, an 18-foot truck is 216 inches long (18 ft x 12 in/ft). The scale drawing shows a length of 3 inches. To find the scale, we set up the ratio of the drawing length to the actual length: 3 in. (drawing) to 216 in. (actual). Simplifying this gives us a scale of 1 in. = 72 in., which translates to 1 in. = 6 ft (since 72 in. ÷ 12 in/ft = 6 ft). Option A (6 in. = 1 ft) is incorrect; it implies a much larger drawing. Option B (1 in. = 15 ft) underestimates the actual size. Option D (15 in. = 1 ft) greatly exaggerates the scale, making the drawing too small.
To determine the scale used for the drawing, we first convert the actual truck length from feet to inches. Since 1 foot equals 12 inches, an 18-foot truck is 216 inches long (18 ft x 12 in/ft). The scale drawing shows a length of 3 inches. To find the scale, we set up the ratio of the drawing length to the actual length: 3 in. (drawing) to 216 in. (actual). Simplifying this gives us a scale of 1 in. = 72 in., which translates to 1 in. = 6 ft (since 72 in. ÷ 12 in/ft = 6 ft). Option A (6 in. = 1 ft) is incorrect; it implies a much larger drawing. Option B (1 in. = 15 ft) underestimates the actual size. Option D (15 in. = 1 ft) greatly exaggerates the scale, making the drawing too small.
Type your answer in the box. You may use numbers, a decimal point (.), and/or a negative sign (-) in your answer.
A company received a shipment of 8 boxes of metal brackets.
• There are 20 metal brackets in each box.
• The total weight of the shipment is 48 pounds.
What is the weight, in pounds, of each metal bracket?
Correct Answer & Rationale
Correct Answer: 0.3
To find the weight of each metal bracket, first calculate the total number of brackets by multiplying the number of boxes (8) by the number of brackets per box (20), resulting in 160 brackets. Next, divide the total weight of the shipment (48 pounds) by the total number of brackets (160). This calculation yields a weight of 0.3 pounds per bracket. Other options may include numbers that misrepresent the division or assume incorrect values for the total brackets or shipment weight. For example, using a weight of 1 pound per bracket would imply only 48 brackets, which contradicts the initial information provided.
To find the weight of each metal bracket, first calculate the total number of brackets by multiplying the number of boxes (8) by the number of brackets per box (20), resulting in 160 brackets. Next, divide the total weight of the shipment (48 pounds) by the total number of brackets (160). This calculation yields a weight of 0.3 pounds per bracket. Other options may include numbers that misrepresent the division or assume incorrect values for the total brackets or shipment weight. For example, using a weight of 1 pound per bracket would imply only 48 brackets, which contradicts the initial information provided.
Fix It Fast is an auto repair shop that employs 10 mechanics. Each day, the shop owner randomly picks 1 mechanic to receive a free lunch. What is the probability the shop owner will pick the same mechanic to receive a free lunch 2 days in a row?
- A. 1\20
- B. 1/100
- C. 1\5
- D. 1\10
Correct Answer & Rationale
Correct Answer: B
To determine the probability of picking the same mechanic two days in a row, we start by recognizing that there are 10 mechanics. On the first day, any mechanic can be chosen, which does not affect the overall probability. On the second day, to pick the same mechanic again, there is only 1 favorable outcome (the chosen mechanic) out of 10 possible mechanics. Thus, the probability of selecting that same mechanic on the second day is 1/10. Since the first day's choice does not influence this, we multiply the probabilities: (1/10) * (1/10) = 1/100. - Option A (1/20) is incorrect as it miscalculates the favorable outcomes. - Option C (1/5) incorrectly assumes a higher likelihood without considering the second day's requirement. - Option D (1/10) only reflects the probability of picking a mechanic on day two, not the two-day scenario.
To determine the probability of picking the same mechanic two days in a row, we start by recognizing that there are 10 mechanics. On the first day, any mechanic can be chosen, which does not affect the overall probability. On the second day, to pick the same mechanic again, there is only 1 favorable outcome (the chosen mechanic) out of 10 possible mechanics. Thus, the probability of selecting that same mechanic on the second day is 1/10. Since the first day's choice does not influence this, we multiply the probabilities: (1/10) * (1/10) = 1/100. - Option A (1/20) is incorrect as it miscalculates the favorable outcomes. - Option C (1/5) incorrectly assumes a higher likelihood without considering the second day's requirement. - Option D (1/10) only reflects the probability of picking a mechanic on day two, not the two-day scenario.