Rahul performed an experiment using 16 windup rubber band single-propeller airplanes. He wound up the propeller a different number of times and recorded the amount of time (in seconds) that the airplane flew for each number of rotations that the propeller was wound. A regression analysis was performed and the partial computer output is given below. The regression equation is
Time = 0.924 + 0.0462 Rotations

Predictor Coef SE Coef T P
Constant 0.9241 0.6413 1.44 0.172
Rotation 0.04625 0.01565 2.96 0.010

S= 0.5426 ; R-Sq= 38.4% ; R -Sq(adj) =34.0%

Required:
What is a 95 percent confidence interval for the slope of the regression line that relates the number of rotations the rubber band is wound and the plane's flight time?

0.0462 ± 2.131(0.01565)

Step-by-step explanation:

The confidence interval for the slope, :

Slope Coefficient = 0.0462

Slope Coefficient ± Margin of error

Margin of Error = Tcritical * standard error

Standard Error (rotation) = 0.01565

We use T distribution, because sample size is small ; n = 16

Tcritical at 95% ; df = n - 1 = 16 - 1 = 15

Tcritical(0.05, 15) = 2.131

Hence,

Margin of Error = 2.131(0.01565)

Slope Coefficient for rotation = 0.0462

95% Confidence interval :

0.0462 ± 2.131(0.01565)

24+x=y

step-by-step explanation:

option (b) is correct

step-by-step explanation:

to find the solution of this system, we need to convert it into reduced row-echelon form. we have,

after applying row operations r2 -> r2/2 and r3 -> r3/3, we get,