2.5 – modeling real world data:. using scatter plots
TRANSCRIPT
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2.5 – Modeling Real World Data:
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2.5 – Modeling Real World Data:
Using Scatter Plots
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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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Ex.1 The table below shows the median selling price of new, privately-owned, one-family houses for some recent years.
Year 1990 1992 1994 1996 1998 2000
Price ($1000)
122.9 121.5 130 140 152.5 169
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Ex.1 The table below shows the median selling price of new, privately-owned, one-family houses for some recent years.
a. Make a scatter plot of the data.
Year 1990 1992 1994 1996 1998 2000
Price ($1000)
122.9 121.5 130 140 152.5 169
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Years Since 1990
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Price
Years Since 1990
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Price
($1000)
Years Since 1990
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Median House Prices
Price
($1000)
Years Since 1990
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Median House Prices
Price
($1000)
0
Years Since 1990
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Median House Prices
Price
($1000)
0 2 4 6 8 10
Years Since 1990
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Median House Prices
Price
($1000)
0 2 4 6 8 10
Years Since 1990
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Median House Prices
Price
($1000)
1200 2 4 6 8
10
Years Since 1990
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Median House Prices
Price
($1000)
140
1200 2 4 6 8
10
Years Since 1990
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Median House Prices
Price
($1000)
140
1200 2 4 6 8
10
Years Since 1990
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Median House Prices
Price
($1000)
140
1200 2 4 6 8
10
Years Since 1990
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Median House Prices
Price
($1000)
140
1200 2 4 6 8
10
Years Since 1990
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Median House Prices
Price
($1000)
140
1200 2 4 6 8
10
Years Since 1990
![Page 20: 2.5 – Modeling Real World Data:. Using Scatter Plots](https://reader035.vdocuments.site/reader035/viewer/2022062301/56649eb25503460f94bb93c4/html5/thumbnails/20.jpg)
Median House Prices
Price
($1000)
140
1200 2 4 6 8
10
Years Since 1990
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Median House Prices
Price
($1000)
140
1200 2 4 6 8
10
Years Since 1990
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Median House Prices
Price
($1000)
140
1200 2 4 6 8
10
Years Since 1990
b. Make a line of fit.
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Median House Prices
Price
($1000)
140
1200 2 4 6 8
10
Years Since 1990
b. Make a line of fit.
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Median House Prices
Price
($1000)
140
1200 2 4 6 8
10
Years Since 1990
b. Make a line of fit.
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c. Find a prediction equation for line of fit.
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c. Find a prediction equation for line of fit.
*Use the best two ordered pairs from b. to find the slope for the line!
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c. 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)
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c. 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
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c. 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
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c. 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
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c. 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
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c. 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
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c. 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
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c. 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
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c. 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)
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c. 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)
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c. 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
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d. Predict the price in 2020.
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d. Predict the price in 2020.
2020 means when x=30 (yrs after 1990)
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d. Predict the price in 2020.
2020 means when x=30 (yrs after 1990)
*Plug 30 in for x!
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d. Predict the price in 2020.
2020 means when x=30 (yrs after 1990)
*Plug 30 in for x!
y = 5.63x + 107.48
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d. 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
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d. 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
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d. 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
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d. 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.