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.
Other Related Questions
The manager of a shipping company plans to use a small truck to ship pipes: The truck has a flatbed trailer with a rectangular surface that is 27 feet long and 8 feet wide. The truck will travel from Atherton to Bakersfield, where some pipes will be delivered, and then on to Castlewood to deliver the remaining pipes. The map shows the roads that connect Atherton. Bakersfield. and Castlewood.
The manager is planning to buy a new truck with better gas mileage. He collected data bout the gas mileage of one of the company's trucks. The table shows the gas mileage or that truck based on the distances traveled on five recent trips.
How many different ways can the truck travel from Atherton to Bakersfield a to Castlewood, using the roads on the map?
- A. 6
- B. 8
- C. 9
- D. 5
Correct Answer & Rationale
Correct Answer: A
To determine the number of different routes from Atherton to Bakersfield and then to Castlewood, we analyze the connections between these locations. If there are 3 distinct paths from Atherton to Bakersfield and 2 distinct paths from Bakersfield to Castlewood, the total number of combinations is found by multiplying the number of options: 3 paths (Atherton to Bakersfield) × 2 paths (Bakersfield to Castlewood) = 6 routes. Options B (8), C (9), and D (5) miscalculate the available paths or overlook the combinations of routes, leading to incorrect totals. Thus, the correct answer accurately reflects the possible travel routes.
To determine the number of different routes from Atherton to Bakersfield and then to Castlewood, we analyze the connections between these locations. If there are 3 distinct paths from Atherton to Bakersfield and 2 distinct paths from Bakersfield to Castlewood, the total number of combinations is found by multiplying the number of options: 3 paths (Atherton to Bakersfield) × 2 paths (Bakersfield to Castlewood) = 6 routes. Options B (8), C (9), and D (5) miscalculate the available paths or overlook the combinations of routes, leading to incorrect totals. Thus, the correct answer accurately reflects the possible travel routes.
Ricardo has two bank accounts. Each month, he will withdraw a certain amount of money from the first account and deposit a different amount of money into the second account. The inequality 8,000 – 200x ? 5,000 + 300x can be solved to find the number of months, x, for which the account has more money than the second account. What is the solution to this inequality?
- A. x ? 6
- B. x ? 30
- C. x ? 30
- D. x ? 6
Correct Answer & Rationale
Correct Answer: D
To solve the inequality \( 8,000 - 200x > 5,000 + 300x \), we first isolate \( x \). Rearranging gives \( 8,000 - 5,000 > 300x + 200x \), simplifying to \( 3,000 > 500x \). Dividing by 500 results in \( x < 6 \). Thus, the solution indicates that for \( x \) to ensure the first account has more money, it must be less than 6 months. Option A incorrectly states \( x \geq 6 \), which contradicts the solution. Options B and C mistakenly suggest \( x \geq 30 \), which is not relevant to the problem.
To solve the inequality \( 8,000 - 200x > 5,000 + 300x \), we first isolate \( x \). Rearranging gives \( 8,000 - 5,000 > 300x + 200x \), simplifying to \( 3,000 > 500x \). Dividing by 500 results in \( x < 6 \). Thus, the solution indicates that for \( x \) to ensure the first account has more money, it must be less than 6 months. Option A incorrectly states \( x \geq 6 \), which contradicts the solution. Options B and C mistakenly suggest \( x \geq 30 \), which is not relevant to the problem.
The graph of the equation y = x^2 + 4x - 5 is shown on the grid. Which statement is true when y = 0?
- A. x= -5 and x=1
- B. x= -2
- C. x= -5 and x = 0
- D. x= -9
Correct Answer & Rationale
Correct Answer: A
To find the values of x when y = 0, we need to solve the equation \(x^2 + 4x - 5 = 0\). Factoring this quadratic gives \((x + 5)(x - 1) = 0\), leading to the solutions \(x = -5\) and \(x = 1\). Option A correctly identifies these solutions. Option B states \(x = -2\), which is not a solution to the equation. Option C suggests \(x = -5\) and \(x = 0\); while it includes one correct solution, \(x = 0\) is incorrect. Option D claims \(x = -9\), which does not satisfy the equation. Thus, only option A accurately reflects the solutions when y = 0.
To find the values of x when y = 0, we need to solve the equation \(x^2 + 4x - 5 = 0\). Factoring this quadratic gives \((x + 5)(x - 1) = 0\), leading to the solutions \(x = -5\) and \(x = 1\). Option A correctly identifies these solutions. Option B states \(x = -2\), which is not a solution to the equation. Option C suggests \(x = -5\) and \(x = 0\); while it includes one correct solution, \(x = 0\) is incorrect. Option D claims \(x = -9\), which does not satisfy the equation. Thus, only option A accurately reflects the solutions when y = 0.
Kelly has a home business making jewellery. It takes 2 hours for her to make each bracelet and 3.5 hours to make each necklace. Next month she plans to spend 140 hours to make jewellery. If she fills a special order for 22 bracelets at the beginning of the mouth and spends the rest of the month making necklaces, how many necklaces can Kelly make in the month
- A. 52
- B. 27
- C. 40
- D. 31
Correct Answer & Rationale
Correct Answer: B
To determine how many necklaces Kelly can make, first calculate the time spent on bracelets. Making 22 bracelets takes 22 x 2 = 44 hours. Subtracting this from her total available time of 140 hours leaves her with 140 - 44 = 96 hours for necklaces. Each necklace takes 3.5 hours, so she can make 96 ÷ 3.5 = 27.43, which rounds down to 27 necklaces since she cannot make a fraction of a necklace. Options A (52), C (40), and D (31) are incorrect because they exceed the available time after accounting for the hours spent on bracelets, indicating miscalculations in time management or misunderstanding of the problem constraints.
To determine how many necklaces Kelly can make, first calculate the time spent on bracelets. Making 22 bracelets takes 22 x 2 = 44 hours. Subtracting this from her total available time of 140 hours leaves her with 140 - 44 = 96 hours for necklaces. Each necklace takes 3.5 hours, so she can make 96 ÷ 3.5 = 27.43, which rounds down to 27 necklaces since she cannot make a fraction of a necklace. Options A (52), C (40), and D (31) are incorrect because they exceed the available time after accounting for the hours spent on bracelets, indicating miscalculations in time management or misunderstanding of the problem constraints.