Applications of Dividing Rational Numbers Quick Check
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Question
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 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 correct response is:
The quotient that describes the rate of change in the squirrel’s height above the ground is -10.48 feet/second.
The correct response 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 need to divide the change in height (45 feet) by the time it takes (5.25 seconds).
Dividing 45 feet by 5.25 seconds, we get approximately 8.57 feet/second.
If we round this to the nearest hundredth, we get 8.57 feet/second.
Therefore, none of the given responses correctly interpret the quotient. The correct answer is not listed.