A helium balloon is at a height of 1,200 ft. after two minutes in flight. The balloon finally pops after seven minutes in flight, at a height of 10,500 ft. What is the average rate of change in height for the balloon over this period of time?(1 point)

Question

Responses

9,300 ft./min.
9,300 ft./min.

2,340 ft./min.
2,340 ft./min.

1,033.33 ft./min.
1,033.33 ft./min.

1,860 ft./min

Answer 1

To calculate the average rate of change in height of the helium balloon over the given period of time, we can use the formula for average rate of change:

\[ \text{Average Rate of Change} = \frac{\text{Change in Height}}{\text{Change in Time}} \]

1. Determine the change in height:

   • Initial height after 2 minutes: \(1,200\) ft
   • Final height after 7 minutes: \(10,500\) ft
   • Change in height = Final height - Initial height \[ = 10,500 , \text{ft} - 1,200 , \text{ft} = 9,300 , \text{ft} \]

2. Determine the change in time:

   • Initial time: \(2\) minutes
   • Final time: \(7\) minutes
   • Change in time = Final time - Initial time \[ = 7 , \text{minutes} - 2 , \text{minutes} = 5 , \text{minutes} \]

3. Calculate the average rate of change: \[ \text{Average Rate of Change} = \frac{9,300 , \text{ft}}{5 , \text{min}} = 1,860 , \text{ft/min} \]

Thus, the average rate of change in height for the balloon over this period of time is \(1,860 , \text{ft/min}\).

So, the correct response is:

1,860 ft./min.