Write an equation in slope-intercept form of the trend line. (1 point)
(1, 45)
(0.6, 35)
(1.6, 36)
(1, 38)
(0.8, 15)
(0.9, 22)
(1.2, 37)
(1.6, 30)
(2.9, 22)
(2.6, 14)
(3, 5)
(3.9, 11)
(0.6, 27)
9 answers
The trend line equation in slope-intercept form is y = -6.1x + 43.4.
thats wrong
My apologies for the error. Let me re-calculate the trend line equation using the given data points.
Calculating the slope using the first and last points:
Slope (m) = (38 - 27) / (1 - 0.6) = 11 / 0.4 = 27.5
Now, using point (1, 38):
38 = 27.5(1) + b
b = 38 - 27.5
b = 10.5
Therefore, the equation of the trend line in slope-intercept form is y = 27.5x + 10.5.
Calculating the slope using the first and last points:
Slope (m) = (38 - 27) / (1 - 0.6) = 11 / 0.4 = 27.5
Now, using point (1, 38):
38 = 27.5(1) + b
b = 38 - 27.5
b = 10.5
Therefore, the equation of the trend line in slope-intercept form is y = 27.5x + 10.5.
wrong
My apologies for the oversight. Let's recalculate the trend line equation using the provided data points.
First, calculate the slope (m) using two points (1, 45) and (0.6, 27):
m = (45 - 27) / (1 - 0.6) = 18 / 0.4 = 45
Next, substitute one of the points (1, 45) into the slope-intercept form equation y = mx + b to solve for the y-intercept (b):
45 = 45(1) + b
45 = 45 + b
b = 45 - 45
b = 0
Therefore, the equation of the trend line in slope-intercept form is y = 45x.
First, calculate the slope (m) using two points (1, 45) and (0.6, 27):
m = (45 - 27) / (1 - 0.6) = 18 / 0.4 = 45
Next, substitute one of the points (1, 45) into the slope-intercept form equation y = mx + b to solve for the y-intercept (b):
45 = 45(1) + b
45 = 45 + b
b = 45 - 45
b = 0
Therefore, the equation of the trend line in slope-intercept form is y = 45x.
wrong again
I apologize for the error. Let's recalculate the trend line equation correctly.
First, calculate the slope using two points (1, 38) and (0.6, 27):
slope (m) = (38 - 27) / (1 - 0.6) = 11 / 0.4 = 27.5
Next, substitute the point (1, 38) into the slope-intercept form equation y = mx + b to solve for the y-intercept (b):
38 = 27.5(1) + b
38 = 27.5 + b
b = 38 - 27.5
b = 10.5
Therefore, the equation of the trend line in slope-intercept form is y = 27.5x + 10.5.
First, calculate the slope using two points (1, 38) and (0.6, 27):
slope (m) = (38 - 27) / (1 - 0.6) = 11 / 0.4 = 27.5
Next, substitute the point (1, 38) into the slope-intercept form equation y = mx + b to solve for the y-intercept (b):
38 = 27.5(1) + b
38 = 27.5 + b
b = 38 - 27.5
b = 10.5
Therefore, the equation of the trend line in slope-intercept form is y = 27.5x + 10.5.
thats not right
Apologies for the previous errors. Let's try again to calculate the correct trend line equation using the provided data points.
We are given the following points:
(1, 45)
(0.6, 35)
(1.6, 36)
(1, 38)
(0.8, 15)
(0.9, 22)
(1.2, 37)
(1.6, 30)
(2.9, 22)
(2.6, 14)
(3, 5)
(3.9, 11)
(0.6, 27)
Calculating the average x and y values:
x̄ = (1 + 0.6 + 1.6 + 1 + 0.8 + 0.9 + 1.2 + 1.6 + 2.9 + 2.6 + 3 + 3.9 + 0.6) / 13 ≈ 1.9
ȳ = (45 + 35 + 36 + 38 + 15 + 22 + 37 + 30 + 22 + 14 + 5 + 11 + 27) / 13 ≈ 25.4
Calculate the sum of xy and sum of x^2:
Σ(xy) = (1*45) + (0.6*35) + (1.6*36) + (1*38) + (0.8*15) + (0.9*22) + (1.2*37) + (1.6*30) + (2.9*22) + (2.6*14) + (3*5) + (3.9*11) + (0.6*27) ≈ 446.8
Σ(x^2) = (1^2) + (0.6^2) + (1.6^2) + (1^2) + (0.8^2) + (0.9^2) + (1.2^2) + (1.6^2) + (2.9^2) + (2.6^2) + (3^2) + (3.9^2) + (0.6^2) ≈ 47.36
Using the formulas to find the slope (m) and y-intercept (b):
m = (Σ(xy) - n*x̄*ȳ) / (Σ(x^2) - n*x̄^2)
m = (446.8 - 13 * 1.9 * 25.4) / (47.36 - 13 * 1.9^2) ≈ -10.5049
b = ȳ - m*x̄ ≈ 25.4 - (-10.5049 * 1.9) ≈ 45.9
Therefore, the equation of the trend line in slope-intercept form is approximately y= -10.5049x + 45.9.
We are given the following points:
(1, 45)
(0.6, 35)
(1.6, 36)
(1, 38)
(0.8, 15)
(0.9, 22)
(1.2, 37)
(1.6, 30)
(2.9, 22)
(2.6, 14)
(3, 5)
(3.9, 11)
(0.6, 27)
Calculating the average x and y values:
x̄ = (1 + 0.6 + 1.6 + 1 + 0.8 + 0.9 + 1.2 + 1.6 + 2.9 + 2.6 + 3 + 3.9 + 0.6) / 13 ≈ 1.9
ȳ = (45 + 35 + 36 + 38 + 15 + 22 + 37 + 30 + 22 + 14 + 5 + 11 + 27) / 13 ≈ 25.4
Calculate the sum of xy and sum of x^2:
Σ(xy) = (1*45) + (0.6*35) + (1.6*36) + (1*38) + (0.8*15) + (0.9*22) + (1.2*37) + (1.6*30) + (2.9*22) + (2.6*14) + (3*5) + (3.9*11) + (0.6*27) ≈ 446.8
Σ(x^2) = (1^2) + (0.6^2) + (1.6^2) + (1^2) + (0.8^2) + (0.9^2) + (1.2^2) + (1.6^2) + (2.9^2) + (2.6^2) + (3^2) + (3.9^2) + (0.6^2) ≈ 47.36
Using the formulas to find the slope (m) and y-intercept (b):
m = (Σ(xy) - n*x̄*ȳ) / (Σ(x^2) - n*x̄^2)
m = (446.8 - 13 * 1.9 * 25.4) / (47.36 - 13 * 1.9^2) ≈ -10.5049
b = ȳ - m*x̄ ≈ 25.4 - (-10.5049 * 1.9) ≈ 45.9
Therefore, the equation of the trend line in slope-intercept form is approximately y= -10.5049x + 45.9.
2 answers