Find the linear regression equation for the transformed data. x=1,2,3,4,5 y=13,19,37,91,253 log y=1.114,1.279,1.568,1.959,2,403
Talja [164]
Answer:
The answer is OPTION (D)log(y)=0.326x+0.687
<h2>
Linear regression:</h2>
It is a linear model, e.g. a model that assumes a linear relationship between the input variables (x) and the single output variable (y)
The Linear regression equation for the transformed data:
We transform the predictor (x) values only. We transform the response (y) values only. We transform both the predictor (x) values and response (y) values.
(1, 13) 1.114
(2, 19) 1.279
(3, 37) 1.568
(4, 91) 1.959
(5, 253) 2.403
X Y Log(y)
1 13 1.114
2 19 1.740
3 37 2.543
4 91 3.381
5 253 4.226
Sum of X = 15
Sum of Y = 8.323
Mean X = 3
Mean Y = 1.6646
Sum of squares (SSX) = 10
Sum of products (SP) = 3.258
Regression Equation = ŷ = bX + a
b = SP/SSX = 3.26/10 = 0.3258
a = MY - bMX = 1.66 - (0.33*3) = 0.6872
ŷ = 0.3258X + 0.6872
The graph is plotted below:
The linear regression equation is log(y)=0.326x+0.687
Learn more about Linear regression equation here:
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These are two questions and two answers.
Question 1) Which of the following polar equations is equivalent to the parametric equations below?
<span>
x=t²
y=2t</span>
Answer: option <span>A.) r = 4cot(theta)csc(theta)
</span>
Explanation:
1) Polar coordinates ⇒ x = r cosθ and y = r sinθ
2) replace x and y in the parametric equations:
r cosθ = t²
r sinθ = 2t
3) work r sinθ = 2t
r sinθ/2 = t
(r sinθ / 2)² = t²
4) equal both expressions for t²
r cos θ = (r sin θ / 2 )²
5) simplify
r cos θ = r² (sin θ)² / 4
4 = r (sinθ)² / cos θ
r = 4 cosθ / (sinθ)²
r = 4 cot θ csc θ ↔ which is the option A.
Question 2) Which polar equation is equivalent to the parametric equations below?
<span>
x=sin(theta)cos(theta)+cos(theta)
y=sin^2(theta)+sin(theta)</span>
Answer: option B) r = sinθ + 1
Explanation:
1) Polar coordinates ⇒ x = r cosθ, and y = r sinθ
2) replace x and y in the parametric equations:
a) r cosθ = sin(θ)cos(θ)+cos(θ)
<span>
b) r sinθ =sin²(θ)+sin(θ)</span>
3) work both equations
a) r cosθ = sin(θ)cos(θ)+cos(θ) ⇒ r cosθ = cosθ [ sin θ + 1] ⇒ r = sinθ + 1
<span>
b) r sinθ =sin²(θ)+sin(θ) ⇒ r sinθ = sinθ [sinθ + 1] ⇒ r = sinθ + 1
</span><span>
</span><span>
</span>Therefore, the answer is r = sinθ + 1 which is the option B.
Answer:
10^4
Step-by-step explanation:
