Triangle JKL with vertices J(1,-1), K(2,3), and L(3,-2) in the line x=4 ANSWER RIGHT NOW BEEN WAITING AN HOUR NOW
1 answer:
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Answer: approximately 24
Step-by-step explanation:
We need to plot a regression line.
So we fit a model using the regression of Y on X, that an equation that predict Y for a given X using:
(Y -mean(Y ))= a(X-meanX)...........1
Where the formular of a is given the attachment.
N= the of individuals = 5
Y = amount of fat
X = time of exercise
mean(X )= sum of all X /N
= 131/5 = 26.2
mean(Y) = sum of all Y/N
= 104/5 = 20.8
a = N(SXY) - (SX)(SY)/ NS(X²) -(SX)²......2
SXY = Sum of Product X and Y
SX= sum of all X
SY = Sum of all Y
S(X²)= sum of all X²
(SX) = square of sum of X
a = -0.478
Hence we substitute into 1
Y-20.8 = -0.478 (X-26.2)
Y -20.8 = -0.478X - 12.524
Y = -0.478X + 33.324 or
Y = 33.324 - 0.478X (model)
When X = 20
Y = 33.324 - 0.478 × 20
Y = 33.324 - 9.56
Y = 23. 764
Y =24(approximately)
Carefully meaning of formula used in attachment to the solution they are the same.
