To determine which option is best for each scenario, let's analyze the total cost for each option and compare them.
1. If you plan to attend the amusement park 3 times this summer:
- Pay per visit: 3 visits x $28/visit = $<<3*28=84>>84.
- Buy a membership: $90 (monthly membership) + 3 visits x $15/visit = $<<90+3*15=135>>135.
In this case, the pay per visit option is cheaper.
2. If you think you will visit the amusement park about 8 times this summer:
- Pay per visit: 8 visits x $28/visit = $<<8*28=224>>224.
- Buy a membership: $90 (monthly membership) + 8 visits x $15/visit = $<<90+8*15=210>>210.
In this case, the monthly membership option is cheaper.
3. If you have budgeted $150 for visiting the park this summer:
- Pay per visit: $150 / $28/visit ≈ 5.36 visits.
- Buy a membership: $90 (monthly membership) + 5 visits x $15/visit = $165.
In this case, the pay per visit option allows you to stay within your budget.
4. To find the breakeven point where both options cost about the same:
Let's assume that the number of visits needed for the options to cost about the same is x:
- Pay per visit: x visits x $28/visit = $28x.
- Buy a membership: $90 (monthly membership) + x visits x $15/visit = $90 + $15x.
- Set both options equal to each other: $28x = $90 + $15x.
- Solve for x: $13x = $90.
- x ≈ $90/$13 ≈ 6.92 visits.
Therefore, approximately 7 visits would be the breakeven point where both options would cost about the same.
An amusement park that you are excited to try is $28 per visit. You have the option to purchase a monthly membership for $90 and then pay $15 for each visit. Put responses in the correct input to answer the question. Select a response, navigate to the desired input and insert the response. Responses can be selected and inserted using the space bar, enter key, left mouse button or touchpad. Responses can also be moved by dragging with a mouse. You have plans to attend 3 times this summer. Which option is best? You think you will visit probably about 8 times this summer. Which option is best? You have budgeted $150 for visiting the park this summer. Which option is best? How many visits would be approximately the breakeven point where both options would cost about the same? Pay per visit Buy a membership 2 5 7 11
5 answers
An amusement park that you are excited to try is $28 per visit. You have the option to purchase a monthly membership for $90 and then pay $15 for each visit. (4 points) Put responses in the correct input to answer the question. Select a response, navigate to the desired input and insert the response. Responses can be selected and inserted using the space bar, enter key, left mouse button or touchpad. Responses can also be moved by dragging with a mouse. You have plans to attend 3 times this summer. Which option is best? You think you will visit probably about 8 times this summer. Which option is best? You have budgeted $150 for visiting the park this summer. Which option is best? How many visits would be approximately the break even point where both options would cost about the same?
To determine which option is best, let's calculate the total cost for each scenario:
1. For 3 visits this summer:
- Pay per visit: 3 visits * $28/visit = $84
- Buy a membership: $90 (monthly membership fee) + 3 visits * $15/visit = $135
In this case, the pay per visit option is cheaper.
2. For 8 visits this summer:
- Pay per visit: 8 visits * $28/visit = $224
- Buy a membership: $90 (monthly membership fee) + 8 visits * $15/visit = $210
In this case, the buy a membership option is cheaper.
3. With a budget of $150:
- Pay per visit: $150 / $28/visit ≈ 5.36 visits (approximately 5 visits)
- Buy a membership: $90 (monthly membership fee) + $150 - $90 budget = $60 for visits / $15/visit = 4 visits
In this case, the buy a membership option allows you to visit 4 times, which is within your budget.
4. To find the break-even point where both options cost about the same:
Set the total cost for both options equal to each other:
- Pay per visit: x visits * $28/visit = Total cost
- Buy a membership: $90 + x visits * $15/visit = Total cost
Setting the equations equal to each other:
x visits * $28/visit = $90 + x visits * $15/visit
Simplifying,
$13x visits = $90
x visits ≈ $90 / $13 ≈ 6.92 visits (approximately 7 visits)
Therefore, approximately 7 visits would be the break-even point where both options would cost about the same.
1. For 3 visits this summer:
- Pay per visit: 3 visits * $28/visit = $84
- Buy a membership: $90 (monthly membership fee) + 3 visits * $15/visit = $135
In this case, the pay per visit option is cheaper.
2. For 8 visits this summer:
- Pay per visit: 8 visits * $28/visit = $224
- Buy a membership: $90 (monthly membership fee) + 8 visits * $15/visit = $210
In this case, the buy a membership option is cheaper.
3. With a budget of $150:
- Pay per visit: $150 / $28/visit ≈ 5.36 visits (approximately 5 visits)
- Buy a membership: $90 (monthly membership fee) + $150 - $90 budget = $60 for visits / $15/visit = 4 visits
In this case, the buy a membership option allows you to visit 4 times, which is within your budget.
4. To find the break-even point where both options cost about the same:
Set the total cost for both options equal to each other:
- Pay per visit: x visits * $28/visit = Total cost
- Buy a membership: $90 + x visits * $15/visit = Total cost
Setting the equations equal to each other:
x visits * $28/visit = $90 + x visits * $15/visit
Simplifying,
$13x visits = $90
x visits ≈ $90 / $13 ≈ 6.92 visits (approximately 7 visits)
Therefore, approximately 7 visits would be the break-even point where both options would cost about the same.
Which method of solving a system of linear equation do you prefer to use? Why do you find this particular method easier than the other two? (2 points)
As an AI language model, I don't have personal preferences. However, I can provide some insights into different methods of solving a system of linear equations.
There are multiple methods to solve a system of linear equations, including elimination/substitution, matrix methods, and graphing. The choice of method depends on the specific situation and personal preferences.
- Elimination/Substitution method: This method involves manipulating the equations to eliminate variables and solve for the remaining variables. It is straightforward and intuitive when dealing with smaller systems, but it can become tedious for larger systems.
- Matrix methods: Matrix methods, such as Gaussian elimination or matrix inversion, involve representing the system of equations as a matrix and applying row operations to solve for the variables. These methods are efficient for larger systems and can be easily implemented using numerical software or calculators.
- Graphing method: This method involves graphing the equations on a coordinate plane and finding the points of intersection as the solution. It is visually intuitive but can be challenging for complex systems or systems with fractional or decimal solutions.
The preference for a specific method may vary depending on personal familiarity, complexity of the system, available tools or technology, and specific requirements of the problem. Some individuals may find elimination/substitution easier for its step-by-step approach, while others may prefer matrix methods for their efficiency and versatility.
In general, it is beneficial to have knowledge and proficiency in multiple methods to choose the most appropriate and efficient method for each specific situation.
There are multiple methods to solve a system of linear equations, including elimination/substitution, matrix methods, and graphing. The choice of method depends on the specific situation and personal preferences.
- Elimination/Substitution method: This method involves manipulating the equations to eliminate variables and solve for the remaining variables. It is straightforward and intuitive when dealing with smaller systems, but it can become tedious for larger systems.
- Matrix methods: Matrix methods, such as Gaussian elimination or matrix inversion, involve representing the system of equations as a matrix and applying row operations to solve for the variables. These methods are efficient for larger systems and can be easily implemented using numerical software or calculators.
- Graphing method: This method involves graphing the equations on a coordinate plane and finding the points of intersection as the solution. It is visually intuitive but can be challenging for complex systems or systems with fractional or decimal solutions.
The preference for a specific method may vary depending on personal familiarity, complexity of the system, available tools or technology, and specific requirements of the problem. Some individuals may find elimination/substitution easier for its step-by-step approach, while others may prefer matrix methods for their efficiency and versatility.
In general, it is beneficial to have knowledge and proficiency in multiple methods to choose the most appropriate and efficient method for each specific situation.