A squirrel has stored its acorns in a hole that is 45 feet from the ground in a tall tree. The squirrel starts on a perch 100 feet above the ground. The squirrel moves from the perch down to its stored acorns in 5.25 seconds. Interpret the quotient to describe the rate of change in the squirrel’s height above the ground. Give your answer to the nearest hundredth.(1 point) Responses The quotient that describes the rate of change in the squirrel’s height above the ground is 10.48 feet/second. The quotient that describes the rate of change in the squirrel’s height above the ground is 10.48 feet/second. The quotient that describes the rate of change in the squirrel’s height above the ground is −10.48 feet/second. The quotient that describes the rate of change in the squirrel’s height above the ground is negative 10.48 feet/second. The quotient that describes the rate of change in the squirrel’s height above the ground is −19.05 feet/second. The quotient that describes the rate of change in the squirrel’s height above the ground is negative 19.05 feet/second. The quotient that describes the rate of change in the squirrel’s height above the ground is 19.05 feet/second.

The quotient that describes the rate of change in the squirrel's height above the ground is $\boxed{10.48}$ feet/second.

The correct interpretation of the quotient is "The quotient that describes the rate of change in the squirrel's height above the ground is 10.48 feet/second."

To find the rate of change in the squirrel's height above the ground, we can divide the change in height by the time it takes.

The squirrel starts on a perch 100 feet above the ground and moves down to the stored acorns, which are 45 feet from the ground. The change in height is 100 - 45 = 55 feet.

The time it takes for the squirrel to make this movement is given as 5.25 seconds.

To find the rate of change, we divide the change in height by the time taken:

Rate of change = Change in height / Time taken

Rate of change = 55 ft / 5.25 sec

Evaluating this expression gives us approximately 10.48 feet/second.

Therefore, the correct answer is "The quotient that describes the rate of change in the squirrel’s height above the ground is 10.48 feet/second."