lashawnford173
contestada

PLEASE HELP ME!!!! I NEED THIS QUICKLY!!
I'LL MARK YOU BRAINLIEST IF YOU ANSWER THIS CORRECTLY!!!
Anika is on the crew to set up rides for the state fair. The crew does most of the setup on the day that the fair arrives at the fairground and then continues to work on finishing the setup for about a week to have the rides ready to go in time for the opening of the fair. The scatter plot shows Anika's setup time on different days and the linear model for the data.

(A). What is the equation of the line, written in slope-intercept form? Show how you determined the equation.


(B). Based on the linear model, predict how long Anika worked on the setup crew on the day the fair arrived at the fairgrounds, Day 0.
Anika worked _______ hours on day 0 <===== Put in decimal

(C). Approximately how much did her setup time decrease per day?
______ Hour and _____ min each day

PLEASE HELP ME I NEED THIS QUICKLY ILL MARK YOU BRAINLIEST IF YOU ANSWER THIS CORRECTLY Anika is on the crew to set up rides for the state fair The crew does mo class=

Respuesta :

Part A: [tex]y=-\frac{5}{4} x+\frac{35}{2}[/tex] is the equation of the line.

Part B: Anika worked 17.5 hours on day 0.

Part C: The setup time decreases with 1 hour and 15 min per day.

Explanation:

Part A: Let us find the equation of the line using the points (2,15) and (6,10)

The equation of the line is given by,

[tex]y-y_{1}=\frac{y_{2}-y_{1}}{x_{2}-x_{1}} \cdot\left(x-x_{1}\right)[/tex]

Substituting, we get,

[tex]y-15=\frac{10-15}{6-2} (x-2)[/tex]

Simplifying, we have,

[tex]y-15=-\frac{5}{4} (x-2)[/tex]

[tex]y-15=-\frac{5}{4} x+\frac{5}{2}[/tex]

Subtracting both sides by 15, we get,

[tex]y=-\frac{5}{4} x+\frac{35}{2}[/tex]

Hence, the equation of the line is [tex]y=-\frac{5}{4} x+\frac{35}{2}[/tex]

Part B: To determine the number of hours Anika worked on day 0, let us substitute x = 0 in the equation of the line.

Thus, we get,

[tex]y=\frac{35}{2}\\y=17.5[/tex]

Hence, Anika worked 17.5 hours on day 0.

Part C: The setup time decrease per day can be determined using the slope.

The formula for slope is given by

[tex]m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}[/tex]

Substituting , we have,

[tex]m=\frac{10-15}{6-2}=-\frac{5}{4}[/tex]

Converting the fraction into hours and minutes, we get,

[tex]-\frac{5}{4}\times60=-75 min[/tex]

Since, 1 hour = 60 min

Subtracting 60 and 75 = 15 min

Thus, the setup time decreases with 1 hour and 15 min per day.

Otras preguntas