Find the first and second moments of the exponential distribution where λ=5.

D. Kelsey graphed the slope as the y-intercept and the y-intercept as the slope.

Step-by-step explanation:

The equation to graph was : y= 3x + 1 where the slope is 3 and the y-intercept is 1

However, Kelsey graph line passes through points (-3, 0)and (0, 3).From these points the equation of the line is;

m=Δy/Δx  where

m=the slope of the graph

Δy= 3-0 =3

Δx= 0--3 =3

m=3/3 = 1

The equation of the line should be;

m=Δy/Δx

1=y-3/x-0

1x= y-3

3+1x= y

y= 1x + 3 ----------where the slope is 1 and y-intercept is 3

So you can see here that in the original equation, the slope m is 3 and y-intercept is 1 as shown in the first attached graph.

While in the Kelsey's graph, the slope is 1 and the y-intercept is 3 as in the second attached graph.

Thus, the correct answer is D : Kelsey graphed the slope,3 as the y-intercept and the y-intercept,1, as the slope.

Answer choice A is incorrect because Kelsey didnot graph the y-intercept, 1 on the x-axis but graphed point (-3,0) on the x-axis.

Answer choice B is incorrect because on Kelsey's equation the y-intercept is 3 and not -3. i.e. y=1x + 3

Answer choice C is incorrect because Kelsey graphed the slope as up 1 right 1 which is okay per the equation , y=1x+3.However, this incorrect because the correct graph has a slope of 3.