Use the table to answer the question.

A grocery store advertises the following cost for oranges.

Number of oranges 1 3 5 9 12
Cost $0.89 $2.67 $4.45 $8.01 $10.68
Can the cost of oranges be modeled by a linear function?

(1 point)
Responses

Yes, because the rate of growth is a variable rate of change.
Yes, because the rate of growth is a variable rate of change.

No, because the rate of growth is a variable rate of change.
No, because the rate of growth is a variable rate of change.

Yes, because the rate of growth is a constant rate of change.
Yes, because the rate of growth is a constant rate of change.

No, because the rate of growth is a constant rate of change.
No, because the rate of growth is a constant rate of change.
Remaining Attempts : 3

1 answer

To determine if the cost of oranges can be modeled by a linear function, we need to check if the rate of change (the cost per number of oranges) is constant.

Let's calculate the cost per orange for different quantities:

  1. From 1 to 3 oranges:

    • Cost for 1 orange = $0.89
    • Cost for 3 oranges = $2.67
    • Change in cost = $2.67 - $0.89 = $1.78 for 2 oranges
    • Average cost per orange = $1.78 / 2 = $0.89
  2. From 3 to 5 oranges:

    • Cost for 3 oranges = $2.67
    • Cost for 5 oranges = $4.45
    • Change in cost = $4.45 - $2.67 = $1.78 for 2 oranges
    • Average cost per orange = $1.78 / 2 = $0.89
  3. From 5 to 9 oranges:

    • Cost for 5 oranges = $4.45
    • Cost for 9 oranges = $8.01
    • Change in cost = $8.01 - $4.45 = $3.56 for 4 oranges
    • Average cost per orange = $3.56 / 4 = $0.89
  4. From 9 to 12 oranges:

    • Cost for 9 oranges = $8.01
    • Cost for 12 oranges = $10.68
    • Change in cost = $10.68 - $8.01 = $2.67 for 3 oranges
    • Average cost per orange = $2.67 / 3 = $0.89

The rate of change is not constant because the change in cost per number of oranges varies (1.78, 3.56, 2.67 are not constant).

Based on this calculation, the correct response would be:

No, because the rate of growth is a variable rate of change.

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