Math Problem Statement

The table below shows the numbers of students, in thousands, who have graduated from high school for the years 1996 to 2005. a) Using x = 0 to correspond to 1996, draw a scatter diagram of the data and write the equation of a linear function that fits the data on P1(1, 2633) and P2(8, 2935). b) Use the model to estimate the expected number of high school graduates in 2010.

Solution

Let's analyze the question and data provided in the image.

The table in the question shows the number of students (in thousands) who graduated from high school each year from 1996 to 2005. Here's a breakdown of the problem requirements:

Problem Details

  1. Data Points (Year and Graduates):

    • The table provides values of xx (year) and yy (number of graduates in thousands).
    • x=x = the year - 1996, so 1996 corresponds to x=0x = 0, 1997 to x=1x = 1, and so on up to 2005 (corresponding to x=9x = 9).
  2. Question Parts:

    • (a) Plot a scatter diagram using the points P1(1,2633)P_1(1, 2633) and P2(8,2935)P_2(8, 2935). Then find the equation of the linear function passing through these points.
    • (b) Use the model obtained in part (a) to estimate the number of high school graduates in 2010.

Solution Steps

Part (a): Linear Equation

  1. Identify the Points:

    • P1(1,2633)P_1(1, 2633): Corresponds to x=1x = 1, y=2633y = 2633.
    • P2(8,2935)P_2(8, 2935): Corresponds to x=8x = 8, y=2935y = 2935.
  2. Calculate the Slope mm: m=y2y1x2x1=2935263381=302743.14m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{2935 - 2633}{8 - 1} = \frac{302}{7} \approx 43.14

  3. Form the Linear Equation: Using the point-slope form yy1=m(xx1)y - y_1 = m(x - x_1): y2633=43.14(x1)y - 2633 = 43.14(x - 1) Expanding and simplifying gives: y=43.14x+2589.86y = 43.14x + 2589.86 So, the equation of the linear function is approximately: y=43.14x+2589.86y = 43.14x + 2589.86

Part (b): Estimate for 2010

  1. Convert Year 2010 to xx:

    • 2010 is 14 years after 1996, so x=14x = 14.
  2. Substitute x=14x = 14 into the Equation: y=43.14(14)+2589.86y = 43.14(14) + 2589.86 Calculating this: y43.1414+2589.86=603.96+2589.863193.82y \approx 43.14 \cdot 14 + 2589.86 = 603.96 + 2589.86 \approx 3193.82 So, the estimated number of graduates in 2010 is approximately 3194 thousand.


Would you like further details on any specific part of the solution?

Additional Questions to Consider

  1. How would the results differ if a different pair of points were chosen?
  2. What would happen if a quadratic model was used instead of a linear one?
  3. How does the accuracy of this linear model change for years further away from the provided data range?
  4. What assumptions are we making by using a linear model for this data?
  5. Could this model be used to estimate graduates for years before 1996?

Tip: For linear models, always verify if a linear trend reasonably fits the data, especially when projecting far from the original range.

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Math Problem Analysis

Mathematical Concepts

Algebra
Linear Equations
Data Analysis
Scatter Plot
Linear Modeling

Formulas

Slope formula: m = (y2 - y1) / (x2 - x1)
Point-slope form of a line: y - y1 = m(x - x1)
Linear equation model: y = mx + b

Theorems

Linear equation theory

Suitable Grade Level

Grades 9-12