A 10-gallon water cooler in an office provides water for the whole department. Each hour, 30 ounces of water are removed from the cooler and drunk by office workers. Write an equation to show how long the water in the cooler will last. (10 gallons is 1,280 ounces.)(1 point)

Question

Responses

30h=10
30 h equals 10

10−30h=0
10 minus 30 h equals 0

1,280−30h=0
1,280 minus 30 h equals 0

1,280+30h=0
1,280 plus 30 h equals 0

Answer 1

To find out how long the water in the cooler will last, we start with the total volume of water in ounces (1,280 ounces) and set up the equation based on the consumption rate.

Each hour, 30 ounces of water are removed from the cooler, so after \( h \) hours, the amount of water consumed is \( 30h \) ounces.

To find out when the cooler will be empty, we want the remaining water to equal zero, which gives us:

\[ 1,280 - 30h = 0 \]

This equation shows that the initial amount of water (1,280 ounces) minus the amount consumed (30 ounces per hour for \( h \) hours) equals zero when the cooler is empty.

Thus, the correct equation is:

\[ 1,280 - 30h = 0 \]

So the response is:

1,280−30h=0 1,280 minus 30 h equals 0