Question: OLYMPIC RECORDS FOR WOMEN'S 200M RACE
Time (seconds) 22.4 22.2 22 21.8 21.6 21.4 21.2
Year 1975 1980 1985 1990 1995 2000 2005 2010 2015 2020 2025 2030
Scatter plot of white ring markers showing Olympic women’s 200m race times over years, with a slightly downward sloping white trend line across the center. position_hint=center
were my calculations for this scatter plot correct?: Calculations
Line of Best Fit:
y=mx+c
Identifying two points on the line:
In 1980, the point was at 22.03
\( (x_1,y_1)=(1980,22.03) \)
In 2024, the point was at 21.83
\( (x_2,y_2)=(2024,21.83) \)
Calculating the gradient:
m=\frac{y_2-y_1}{x_2-x_1}
m=\frac{21.83-22.03}{2024-1980}
m= \frac{-0.2}{44}
m= -0.0045
Using the point-slope form to write the linear equation:
y-y_1=m(x-x_1)
y-22.03=-0.0045(x-1980)
y-22.03=-0.0045x+8.91
y=-0.0045x+8.91+22.03
y=-0.0045x+30.94
\( \therefore \) the linear equation for Graph 1 is \( y=-0.0045x+30.94 \)
1. **State the problem:** We want to verify if the calculations for the line of best fit for Olympic women's 200m race times over years are correct.
2. **Formula used:** The line of best fit is a linear equation \( y=mx+c \), where \( m \) is the slope (gradient) and \( c \) is the y-intercept.
3. **Identify points:** Two points on the line are given:
\( (x_1,y_1) = (1980, 22.03) \)
\( (x_2,y_2) = (2024, 21.83) \)
4. **Calculate the slope \( m \):**
\[ m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{21.83 - 22.03}{2024 - 1980} = \frac{-0.2}{44} = -0.00454545... \approx -0.0045 \]
5. **Use point-slope form:**
\[ y - y_1 = m(x - x_1) \]
Substitute \( m = -0.0045 \), \( x_1 = 1980 \), \( y_1 = 22.03 \):
\[ y - 22.03 = -0.0045(x - 1980) \]
6. **Expand and simplify:**
\[ y - 22.03 = -0.0045x + 0.0045 \times 1980 \]
Calculate \( 0.0045 \times 1980 = 8.91 \):
\[ y - 22.03 = -0.0045x + 8.91 \]
7. **Add 22.03 to both sides:**
\[ y = -0.0045x + 8.91 + 22.03 \]
\[ y = -0.0045x + 30.94 \]
8. **Conclusion:** The calculations are correct and the linear equation for the line of best fit is \( y = -0.0045x + 30.94 \).