Answer :
To determine the linear function that models the cost [tex]\( c \)[/tex] in terms of the number of students [tex]\( s \)[/tex], we need to consider the following:
- The fixed, nonrefundable deposit of [tex]$50. - The additional charge of $[/tex]4.50 for each student.
Given this information, the total cost [tex]\( c \)[/tex] can be modeled by the linear function:
[tex]\[ c = 4.5s + 50 \][/tex]
This equation indicates that for every student [tex]\( s \)[/tex], the cost increases by [tex]$4.50, starting from a base cost of $[/tex]50.
Next, let's analyze the given statements one by one:
1. The linear model is [tex]\( f(s) = 4.5s + 50 \)[/tex].
- Correct. This matches our derived linear function.
2. The linear model is [tex]\( f(c) = 54.5c + 24 \)[/tex].
- Incorrect. This equation does not accurately represent the cost function in terms of either the cost or the number of students.
3. The domain is [tex]\( \{x \mid 0 \leq x \leq 24\} \)[/tex].
- Correct. This makes sense because the number of students [tex]\( s \)[/tex] can vary from 0 to 24.
4. The range is [tex]\( \{y \mid 50 \leq y \leq 158\} \)[/tex].
- Correct. When there are no students ( [tex]\( s = 0 \)[/tex] ), the minimum cost is [tex]$50. For the maximum number of students ( \( s = 24 \) ), the cost is \( 4.5 \times 24 + 50 = 158 \). So, the range of the total cost \( y \) is from $[/tex]50 to $158.
5. The graph is continuous.
- Correct. Costs can vary continuously with the number of students, even though practically the number of students is a whole number. Therefore, it is appropriate to consider the graph as continuous in this context.
In summary, the correct statements are:
- The linear model is [tex]\( f(s) = 4.5s + 50 \)[/tex].
- The domain is [tex]\( \{x \mid 0 \leq x \leq 24\} \)[/tex].
- The range is [tex]\( \{y \mid 50 \leq y \leq 158\} \)[/tex].
- The graph is continuous.
Thus, the correct answers to the given statements are:
- True for the first statement.
- False for the second statement.
- True for the third statement.
- True for the fourth statement.
- True for the fifth statement.
- The fixed, nonrefundable deposit of [tex]$50. - The additional charge of $[/tex]4.50 for each student.
Given this information, the total cost [tex]\( c \)[/tex] can be modeled by the linear function:
[tex]\[ c = 4.5s + 50 \][/tex]
This equation indicates that for every student [tex]\( s \)[/tex], the cost increases by [tex]$4.50, starting from a base cost of $[/tex]50.
Next, let's analyze the given statements one by one:
1. The linear model is [tex]\( f(s) = 4.5s + 50 \)[/tex].
- Correct. This matches our derived linear function.
2. The linear model is [tex]\( f(c) = 54.5c + 24 \)[/tex].
- Incorrect. This equation does not accurately represent the cost function in terms of either the cost or the number of students.
3. The domain is [tex]\( \{x \mid 0 \leq x \leq 24\} \)[/tex].
- Correct. This makes sense because the number of students [tex]\( s \)[/tex] can vary from 0 to 24.
4. The range is [tex]\( \{y \mid 50 \leq y \leq 158\} \)[/tex].
- Correct. When there are no students ( [tex]\( s = 0 \)[/tex] ), the minimum cost is [tex]$50. For the maximum number of students ( \( s = 24 \) ), the cost is \( 4.5 \times 24 + 50 = 158 \). So, the range of the total cost \( y \) is from $[/tex]50 to $158.
5. The graph is continuous.
- Correct. Costs can vary continuously with the number of students, even though practically the number of students is a whole number. Therefore, it is appropriate to consider the graph as continuous in this context.
In summary, the correct statements are:
- The linear model is [tex]\( f(s) = 4.5s + 50 \)[/tex].
- The domain is [tex]\( \{x \mid 0 \leq x \leq 24\} \)[/tex].
- The range is [tex]\( \{y \mid 50 \leq y \leq 158\} \)[/tex].
- The graph is continuous.
Thus, the correct answers to the given statements are:
- True for the first statement.
- False for the second statement.
- True for the third statement.
- True for the fourth statement.
- True for the fifth statement.