Responses
$425.00
$1,265.00
$50.00
$375.00
7 answers
To solve this problem, we can set up an equation where:
The total cost consists of a flat fee plus the cost per ticket for each student. The equation will be:
\[ y = 50 + 15x \]
Now, substitute \( x = 25 \) into the equation:
\[ y = 50 + 15(25) \] \[ y = 50 + 375 \] \[ y = 425 \]
Therefore, the total cost for a class of 25 students to see The Nutcracker is $425.00.
The answer is $425.00.
A discrete data constraint is one that can take on distinct, separate values, typically counts or whole numbers. Let's evaluate the options:
Average weight of all the newborns in a hospital ward: This represents continuous data, as weight can take any value and is not restricted to whole numbers.
Average score on a test: This also can be considered continuous, as test scores may have decimal points (e.g., 87.5).
Number of tests passed: This represents discrete data because you cannot pass a fraction of a test. The values are whole numbers (0, 1, 2, etc.).
Distance between points: This represents continuous data as distance can take on any value (including decimals).
Out of these options, "number of tests passed" represents a discrete data constraint with viable solutions.
To determine which of the listed options could use both intercepts to produce viable solutions, we need to understand the context of two-dimensional data in which each variable can be evaluated as an intercept on a graph.
(age, weight): Age and weight are both continuous variables and can theoretically be plotted on a graph. However, age cannot be negative, so the y-intercept (weight) might not reflect a viable solution if the graph does not consider that age starts from zero.
(years of experience, starting salary): Years of experience cannot be negative (you can't have negative years of experience), and starting salary will likely also be non-negative. This means we can find a point where years of experience are zero (the y-intercept) when starting salary is present. However, both variables are typically non-negative, impacting the usefulness of both intercepts.
(number of nominations, number of awards): Both nominations and awards are discrete, non-negative integers. There is a natural scenario where both can have interceptions at zero (0 nominations and 0 awards), and they can be plotted effectively.
(temperature, number of car rentals): Temperature is continuous and can be negative, while the number of car rentals is non-negative. This could lead to a scenario where temperature intercepts at negative values, complicating interpretations regarding viable solutions.
The strongest candidate for using both intercepts effectively with viable solutions is (number of nominations, number of awards), as both can be graphed in a way that meaningful intercepts can be found.
To find a viable solution for the number of robotics kits (y) and chemistry sets (x) Cindy can purchase with her $50.00, we can set up the equation based on their costs:
Given:
Cindy's total winnings can be expressed in the equation: \[ 10y + 8x = 50 \]
Now, let's evaluate the options to see which ones satisfy this equation:
(5, 1): \[ 10(5) + 8(1) = 50 + 8 = 58 \quad (\text{not viable}) \]
(3, 2.6): \[ 10(3) + 8(2.6) = 30 + 20.8 = 50.8 \quad (\text{not viable}) \]
(−5, 9): \[ 10(-5) + 8(9) = -50 + 72 = 22 \quad (\text{not viable}) \]
(0, 5): \[ 10(0) + 8(5) = 0 + 40 = 40 \quad (\text{not valid; does not use entire winnings}) \]
None of the mentioned options satisfy the equation, as they either exceed the total winnings or do not spend the entire amount of $50.
However, to find a viable solution, we can slightly redefine our approach. Let's see if 0 and 5 can work depending on the context of the expenses: Cindy could buy 5 chemistry sets and not purchase the robotics kits, thus not using her entire winnings but maximizing the chemistry sets.
Given the constraints of the problem as worded, none of the provided options result in an exact use of the winnings. Therefore, if it strictly requires spending all winnings per an example (something ideally rounded in logical integer values), then we'd need options correctly representing a combination within those limits (sum of 50).
