Based on the table from the previous question, determine a mathematical model that describes the amount of candy Shelly has. Define the variables you use.

# of days: 0, 1, 2, 3, 4, 5, 6, 7
# of candy: 160, 80, 40, 20, 10, 5, 2.5, 1.25

1 answer

To develop a mathematical model that describes the amount of candy Shelly has over time, we can identify the amount of candy she has on each day and look for a pattern.

Given the data points:

  • Day 0: 160 candies
  • Day 1: 80 candies
  • Day 2: 40 candies
  • Day 3: 20 candies
  • Day 4: 10 candies
  • Day 5: 5 candies
  • Day 6: 2.5 candies
  • Day 7: 1.25 candies

Observing the Pattern

The amount of candy appears to decrease by a factor of 2 each day:

  • From Day 0 to Day 1, the candy amount halves: \( 160 \div 2 = 80 \)
  • From Day 1 to Day 2, it again halves: \( 80 \div 2 = 40 \)
  • This pattern continues down to Day 7.

Deriving the Mathematical Model

From the observed pattern, we can derive a model in the form of an exponential decay function.

We can define our variables:

  • Let \( C(d) \) be the amount of candy Shelly has on day \( d \).
  • Let \( d \) be the number of days since Day 0.

Model Formulation

The amount of candy can be mathematically expressed as:

\[ C(d) = C_0 \times \left( \frac{1}{2} \right)^d \]

where:

  • \( C_0 = 160 \) is the initial amount of candy (at \( d = 0 \)).
  • \( d \) is the number of days.

Final Expression

Plugging in the value for \( C_0 \):

\[ C(d) = 160 \times \left( \frac{1}{2} \right)^d \]

This model accurately describes the amount of candy Shelly has on each day from day 0 to day 7.

Verification

To ensure this model is accurate, we can verify by substituting in the days:

  • \( C(0) = 160 \times (1) = 160 \)
  • \( C(1) = 160 \times \left( \frac{1}{2} \right)^1 = 160 \times 0.5 = 80 \)
  • \( C(2) = 160 \times \left( \frac{1}{2} \right)^2 = 160 \times 0.25 = 40 \)
  • \( C(3) = 160 \times \left( \frac{1}{2} \right)^3 = 160 \times 0.125 = 20 \)
  • \( C(4) = 160 \times \left( \frac{1}{2} \right)^4 = 160 \times 0.0625 = 10 \)
  • \( C(5) = 160 \times \left( \frac{1}{2} \right)^5 = 160 \times 0.03125 = 5 \)
  • \( C(6) = 160 \times \left( \frac{1}{2} \right)^6 = 160 \times 0.015625 = 2.5 \)
  • \( C(7) = 160 \times \left( \frac{1}{2} \right)^7 = 160 \times 0.0078125 = 1.25 \)

All values match the original data. Thus, the model is confirmed to be correct.