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

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

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

1,860 ft./min.
1,860 ft./min.

9,300 ft./min.
9,300 ft./min.
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Answer 1

To find the average rate of change in height for the balloon over the time period, you can use the formula:

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

First, calculate the change in height:

\[ \text{Change in Height} = \text{Final Height} - \text{Initial Height} = 10,500 \text{ ft} - 1,200 \text{ ft} = 9,300 \text{ ft} \]

Next, calculate the change in time:

\[ \text{Change in Time} = \text{Final Time} - \text{Initial Time} = 7 \text{ min} - 2 \text{ min} = 5 \text{ min} \]

Now, plug these values into the average rate of change formula:

\[ \text{Average Rate of Change} = \frac{9,300 \text{ ft}}{5 \text{ min}} = 1,860 \text{ ft/min} \]

Thus, the average rate of change in height for the balloon over this period of time is 1,860 ft/min.

So, the correct response is:

1,860 ft./min.