Math  /  Data & Statistics

Question2. Samantha has been studying the patterns in her school's cafeteria food prices over the past few weeks. She collects data on the cost of lunch each day: - Day 1: Pizza costs \3.50.Day2:Pizzacosts$4.00.Day3:Pizzacosts$4.50.Day4:Pizzacosts3.50. - Day 2: Pizza costs \$4.00. - Day 3: Pizza costs \$4.50. - Day 4: Pizza costs \5.00 5.00.
Question: Predict the price of pizza on Day 7 .

Studdy Solution

STEP 1

1. The price of pizza changes linearly over the days.
2. The pattern of price increase is consistent and can be modeled using a linear equation.

STEP 2

1. Identify the pattern of price increase.
2. Develop a linear equation to model the price change.
3. Use the equation to predict the price on Day 7.

STEP 3

List the prices and corresponding days:
- Day 1: P1=3.50 P_1 = 3.50 - Day 2: P2=4.00 P_2 = 4.00 - Day 3: P3=4.50 P_3 = 4.50 - Day 4: P4=5.00 P_4 = 5.00

STEP 4

Calculate the daily increase in price.
From Day 1 to Day 2: 4.003.50=0.50 4.00 - 3.50 = 0.50
From Day 2 to Day 3: 4.504.00=0.50 4.50 - 4.00 = 0.50
From Day 3 to Day 4: 5.004.50=0.50 5.00 - 4.50 = 0.50
The daily increase is consistent at 0.50 0.50 .

STEP 5

Develop the linear equation for the price P P on day n n .
The pattern shows a linear increase, so we use the equation of a line:
P(n)=P1+(n1)×increase per day P(n) = P_1 + (n - 1) \times \text{increase per day}
Substitute the known values:
P(n)=3.50+(n1)×0.50 P(n) = 3.50 + (n - 1) \times 0.50

STEP 6

Use the equation to predict the price on Day 7.
Substitute n=7 n = 7 into the equation:
P(7)=3.50+(71)×0.50 P(7) = 3.50 + (7 - 1) \times 0.50
P(7)=3.50+6×0.50 P(7) = 3.50 + 6 \times 0.50
P(7)=3.50+3.00 P(7) = 3.50 + 3.00
P(7)=6.50 P(7) = 6.50
The predicted price of pizza on Day 7 is:
6.50 \boxed{6.50}

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