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:

nothjing

step-by-step explanation:

la parábola es el lugar

geométrico en el cual los puntos tienen una distancia idéntica, equidistantes,

de un punto denominado foco, f, y una recta denominada directriz

el radio vector de una

parábola es el segmento que existe entre un punto cualquiera y el foco.

por la ecuación se observa

que la parábola abre hacia la derecha, por lo cual podemos usar las siguientes fórmulas

la ecuación ordinaria de la

parábola es y^2 = 4px, donde p es la longitud desde el vértice de la parábola al

foco.

la fórmula de la directriz

es x + p = 0

resolviendo:

y^2-9x = 0

y^2 = 9x

4p = 9, entonces p = 9/4

entonces el foco tendrá la

coordenada f(9/4,0)

la directriz vendrá

expresada por

x + p = 0

x = -p

x = -9/4

el problema indica que la

ordenada es igual a 6, entonces

(6)^2=9x

x = 36/9

x = 4

entonces el punto al cual

hay que obtener la distancia será p(4,6)

calculando se tendrá: pf = 

pf = 6,25 u

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step-by-step explanation: