The linear regression equation for a data set is y=1.6x-4.1. If the actual value at x=3 is 0.9, what is the residual value at x=3?

Residual is the difference between the predicted value and the actual value of the response variable

The residual at x = 3 is 0.2

The linear regression is given as:

The predicted value at x = 3 is calculated by substituting 3 for x in the linear regression

So, we have:

The actual value is given as 0.9.

So, the residual is calculated using:

This gives

Hence, the residual at x = 3 is 0.2

Read more about residuals at:

link

0.2

Step-by-step explanation:

Given the regression model :

y = 1.6x - 4.1

At x = 3

Predicted value = 1.6(3) - 4.1 = 0.7

If actual value at x = 3 is 0.9

The residual value at x = 3 ;

Residual value is the difference between the actual and predicted value :

Actual at x = 3 minus predicted at x = 3

Hence, Residual = 0.9 - 0.7 = 0.2

the correct functions are

h(x) = 3x + 2  

g(x) = 2^x

step-by-step explanation:

i proceed to calculate the different values ​​of k (x)

case 1)  k(x) = g(x) ∘ h(x)

g(x) ∘ h(x)=g(h(x))

k(x) = [2]^[3x + > is the option  4. k(x) = 2^(3x + 2 )

case 2)  k(x) = g(x) + h(x)

k(x) = [2^x]+[3x + 2]=2^x+3x+> is the option  1.) k(x) = 2^x + 3x + 2

case 3)k(x) = h(x) ÷ g(x)

k(x) =  [3x + 2]/[2^x]=3x/(2^x)+2/(2^x)=3x*(2^-x)+2^(1-x)

is the option  3.)   k(x) = (3x)2^(-x) + 2^(-x + 1 )

case 4)k(x) = g(x) - h(x)

k(x) =  [2^x]-[3x + 2]=2^x-3x-> is the option  2.) k(x) = 2^x - 3x - 2 

case 5)  k(x) = g(x) × h(x)

k(x) = [2^x]*[3x + > is the option  5.) k(x) = 2^x(3x + 2)

case 6)k(x) = h(x) ∘ g(x)

h(x) ∘ g(x)=h(g(x))

k(x)=3*[2^x]+> is the option  6. k(x) = 3(2^x) +2