To rent a certain meeting room, a college charges a reservation fee of $14 and an additional fee of $6 per hour. The chemistry club wants to spend less than $80 on renting the room. What are the possible numbers of hours the chemistry club could rent the meeting room?
Use t for the number of hours.
The total cost to rent the room for t hours is given by the equation:
Cost(t) = 14 + 6t
To find the possible numbers of hours the chemistry club could rent the meeting room, we need to find values of t that make the total cost less than $80.
Therefore, we solve the inequality:
14 + 6t < 80
Subtracting 14 from both sides and simplifying, we have:
6t < 66
Dividing both sides by 6, we get:
t < 11
The possible numbers of hours the chemistry club could rent the meeting room are any whole number less than 11.
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To find the possible numbers of hours the chemistry club could rent the meeting room, we need to set up an inequality based on the given information.
The reservation fee is $14, and the additional fee is $6 per hour. So, the total cost for renting the room can be expressed as:
Total cost = Reservation fee + Additional fee per hour * Number of hours
Total cost = $14 + $6t
Since the chemistry club wants to spend less than $80, we can form the inequality:
Total cost < $80
$14 + $6t < $80
Now, we can solve this inequality to find the range of possible values for t (number of hours).
Subtracting $14 from both sides of the inequality:
$6t < $80 - $14
$6t < $66
Dividing both sides of the inequality by 6:
t < $66/$6
t < 11
Therefore, the possible numbers of hours the chemistry club could rent the meeting room are any whole number less than 11.
To determine the possible number of hours the chemistry club could rent the meeting room, we need to find the values of t that satisfy the given condition of spending less than $80.
We can create an inequality to represent this situation:
Reservation fee + Additional fee per hour * Number of hours < $80
The reservation fee is $14 and the additional fee per hour is $6, so the inequality becomes:
14 + 6t < 80
To solve this inequality for t, we can isolate the variable t by subtracting 14 from both sides:
6t < 80 - 14
6t < 66
Dividing both sides of the inequality by 6:
t < 66/6
t < 11
Therefore, the possible numbers of hours the chemistry club could rent the meeting room are any values of t that are less than 11 (since the club wants to spend less than $80). The range of possible values for t is 0 < t < 11.