Find the residual values, and use the graphing calculator tool to make a residual plot. A 4-column table with 5 rows. The first column is labeled x with entries 1, 2, 3, 4, 5. The second column is labeled given with entries negative 2.7, negative 0.9, 1.1, 3.2, 5.4. The third column is labeled predicted with entries negative 2.84, negative 0.81, 1.22, 3.25, 5.28. The fourth column is labeled residual value with all entries blank. Does the residual plot show that the line of best fit is appropriate for the data? No, the points are in a curved pattern. No, the points are evenly distributed about the x-axis. Yes, the points are in a linear pattern. Yes, the points have no pattern.

Solution-

We know that,

Residual value = Given value - Predicted value

The table for residual values is shown below,

Plotting a graph, by taking the residual values on ordinate and values of given x on abscissa, a random pattern is obtained where the points are evenly distributed about x-axis.

We know that,

If the points in a residual plot are randomly dispersed around the horizontal or x-axis, a linear regression model is appropriate for the data. Otherwise, a non-linear model is more appropriate.

As, in this case the points are distributed randomly around x-axis, so the residual plot show that the line of regression is best fit for the data set.

Hope this helps!

Step-by-step explanation:

∠1≅∠2 by the alternate exterior angles theorem.

step-by-step explanation:

given, a ∥ b and ∠1 ≅ ∠3 .we have to prove that e ∥ f

we know that ∠1≅∠3 and that a || b because they are given. we see that by the alternate exterior angles theorem. therefore, ∠2≅∠3 by the transitive property. so, we can conclude that e || f by the converse alternate exterior angles theorem.

we have to fill the missing statement.

transitivity property states that if a = b and b = c, then a = c.

now, given ∠1≅∠3 and by transitivity property ∠2≅∠3 .

hence, to apply transitivity property one angle must be common which is not in result after applying this property which is ∠1.

the only options in which ∠1 is present are ∠1 and ∠2, ∠1 and ∠4

∠1 and ∠4 is not possible ∵ after applying transitivity we didn't get ∠4.

hence, the missing statement is ∠1≅∠2.

so, ∠1≅∠2 by the alternate exterior angles theorem.

where is abdul's work? don't see that anywhere posted : /

step-by-step explanation: