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2.5 – Modeling Real World Data:

2.5 – Modeling Real World Data:. 2.5 – Modeling Real World Data:. Using Scatter Plots. Ex.1 The table below shows the median selling price of new, privately-owned, one-family houses for some recent years.

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2.5 – Modeling Real World Data:

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  1. 2.5 – Modeling Real World Data:

  2. 2.5 – Modeling Real World Data: Using Scatter Plots

  3. Ex.1 The table below shows the median selling price of new, privately-owned, one-family houses for some recent years.

  4. Ex.1 The table below shows the median selling price of new, privately-owned, one-family houses for some recent years.

  5. Ex.1 The table below shows the median selling price of new, privately-owned, one-family houses for some recent years. • Make a scatter plot of the data.

  6. Years Since 1990

  7. Price Years Since 1990

  8. Price ($1000) Years Since 1990

  9. Median House Prices Price ($1000) Years Since 1990

  10. Median House Prices Price ($1000) 0 Years Since 1990

  11. Median House Prices Price ($1000) 0 2 4 6 8 10 Years Since 1990

  12. Median House Prices Price ($1000) 0 2 4 6 8 10 Years Since 1990

  13. Median House Prices Price ($1000) 120 0 2 4 6 8 10 Years Since 1990

  14. Median House Prices Price ($1000) 140 120 0 2 4 6 8 10 Years Since 1990

  15. Median House Prices Price ($1000) 140 120 0 2 4 6 8 10 Years Since 1990

  16. Median House Prices Price ($1000) 140 120 0 2 4 6 8 10 Years Since 1990

  17. Median House Prices Price ($1000) 140 120 0 2 4 6 8 10 Years Since 1990

  18. Median House Prices Price ($1000) 140 120 0 2 4 6 8 10 Years Since 1990

  19. Median House Prices Price ($1000) 140 120 0 2 4 6 8 10 Years Since 1990

  20. Median House Prices Price ($1000) 140 120 0 2 4 6 8 10 Years Since 1990

  21. Median House Prices Price ($1000) 140 120 0 2 4 6 8 10 Years Since 1990 b. Make a line of fit.

  22. Median House Prices Price ($1000) 140 120 0 2 4 6 8 10 Years Since 1990 b. Make a line of fit.

  23. Median House Prices Price ($1000) 140 120 0 2 4 6 8 10 Years Since 1990 b. Make a line of fit.

  24. Find a prediction equation for line of fit.

  25. Find a prediction equation for line of fit. *Use the best two ordered pairs from b. to find the slope for the line!

  26. Find a prediction equation for line of fit. *Use the best two ordered pairs from b. to find the slope for the line! (4, 130) and (8, 152.5)

  27. Find a prediction equation for line of fit. *Use the best two ordered pairs from b. to find the slope for the line! (4, 130) and (8, 152.5) m = y2 – y1 x2 - x1

  28. Find a prediction equation for line of fit. *Use the best two ordered pairs from b. to find the slope for the line! (4, 130) and (8, 152.5) m = y2 – y1 = 152.2 – 130 x2 - x1 8 – 4

  29. Find a prediction equation for line of fit. *Use the best two ordered pairs from b. to find the slope for the line! (4, 130) and (8, 152.5) m = y2 – y1 = 152.2 – 130 = 22.5 x2 - x1 8 – 4 4

  30. Find a prediction equation for line of fit. *Use the best two ordered pairs from b. to find the slope for the line! (4, 130) and (8, 152.5) m = y2 – y1 = 152.2 – 130 = 22.5≈ 5.63 x2 - x1 8 – 4 4

  31. Find a prediction equation for line of fit. *Use the best two ordered pairs from b. to find the slope for the line! (4, 130) and (8, 152.5) m = y2 – y1 = 152.2 – 130 = 22.5≈ 5.63 x2 - x1 8 – 4 4 *So use x1 = 4

  32. Find a prediction equation for line of fit. *Use the best two ordered pairs from b. to find the slope for the line! (4, 130) and (8, 152.5) m = y2 – y1 = 152.2 – 130 = 22.5≈ 5.63 x2 - x1 8 – 4 4 *So use x1 = 4, y1 = 130

  33. Find a prediction equation for line of fit. *Use the best two ordered pairs from b. to find the slope for the line! (4, 130) and (8, 152.5) m = y2 – y1 = 152.2 – 130 = 22.5≈ 5.63 x2 - x1 8 – 4 4 *So use x1 = 4, y1 = 130, and m≈ 5.63

  34. Find a prediction equation for line of fit. *Use the best two ordered pairs from b. to find the slope for the line! (4, 130) and (8, 152.5) m = y2 – y1 = 152.2 – 130 = 22.5≈ 5.63 x2 - x1 8 – 4 4 *So use x1 = 4, y1 = 130, and m≈ 5.63 y – y1 = m(x – x1)

  35. Find a prediction equation for line of fit. *Use the best two ordered pairs from b. to find the slope for the line! (4, 130) and (8, 152.5) m = y2 – y1 = 152.2 – 130 = 22.5≈ 5.63 x2 - x1 8 – 4 4 *So use x1 = 4, y1 = 130, and m≈ 5.63 y – y1 = m(x – x1)

  36. Find a prediction equation for line of fit. *Use the best two ordered pairs from b. to find the slope for the line! (4, 130) and (8, 152.5) m = y2 – y1 = 152.2 – 130 = 22.5≈ 5.63 x2 - x1 8 – 4 4 *So use x1 = 4, y1 = 130, and m≈ 5.63 y – y1 = m(x – x1) y – 130 = 5.63(x – 4) y – 130 = 5.63(x) – 5.63(4) y – 130 = 5.63x – 22.52 y = 5.63x + 107.48

  37. Predict the price in 2020.

  38. Predict the price in 2020. 2020 means when x=30 (yrs after 1990)

  39. Predict the price in 2020. 2020 means when x=30 (yrs after 1990) *Plug 30 in for x!

  40. Predict the price in 2020. 2020 means when x=30 (yrs after 1990) *Plug 30 in for x! y = 5.63x + 107.48

  41. Predict the price in 2020. 2020 means when x=30 (yrs after 1990) *Plug 30 in for x! y = 5.63x + 107.48 y = 5.63(30) + 107.48

  42. Predict the price in 2020. 2020 means when x=30 (yrs after 1990) *Plug 30 in for x! y = 5.63x + 107.48 y = 5.63(30) + 107.48 y = 168.9 + 107.48

  43. Predict the price in 2020. 2020 means when x=30 (yrs after 1990) *Plug 30 in for x! y = 5.63x + 107.48 y = 5.63(30) + 107.48 y = 168.9 + 107.48 y = 276.38

  44. Predict the price in 2020. 2020 means when x=30 (yrs after 1990) *Plug 30 in for x! y = 5.63x + 107.48 y = 5.63(30) + 107.48 y = 168.9 + 107.48 y = 276.38 So, in 2020 the price will be $276,380.

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