Question 747·Hard·Math

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The scatterplot shows the relationship between $d$ and $t$ for data set P, where $d$ is the number of prototypes a team has already assembled and $t$ is the time (in hours) it takes to assemble the next prototype. A line of best fit is shown.

A new data set, P2, is created by transforming each point $(d,t)$ in data set P to a point $(U,V)$ using the rules

U=d+1

and

V=3(t-5).

Which choice is the $V$ -intercept of a line of best fit for data set P2?

With a line of best fit already drawn, use two marked points on the line to find its slope; the scattered dots need not sit exactly on the model. A new vertical intercept occurs where the new input is zero, so translate that zero back to the old input before making a prediction. Apply the new output rule last, or you'll find an intercept for the wrong input.

Hints

Hint 1
The open circles lie on the fitted line, so use them to find how much $t$ changes when $d$ increases. The solid dots are measurements and don't have to lie exactly on that line.
Hint 2
A vertical intercept has a horizontal coordinate of zero. Here the new horizontal coordinate is $U$ , not $d$ . What value of $d$ makes $U=d+1$ equal zero?
Hint 3
Start at a marked point on the line and use its slope to predict $t$ at that old $d$ value. Then follow the parentheses in $V=3(t-5)$ : subtract $5$ before multiplying by $3$ .

Step-by-step

Find the old prediction at the new intercept
Step 1
Find the fitted line's slope
The open circles $(2,15)$ and $(8,9)$ are on the fitted line, unlike the solid measurement dots. The slope is the change in $t$ divided by the change in $d$ . Type $\frac{9-15}{8-2}$ in Desmos; it shows $-1$ . So the fitted line predicts $1$ fewer hour for each increase of $1$ in $d$ .
Step 2
Find the old input where the new input is zero
The $V$ -intercept is where $U=0$ , so $U=d+1$ becomes $0=d+1$ . Type $d+1\sim0$ in Desmos; $\sim$ tells it to find the value that makes both sides match. Under PARAMETERS, Desmos shows $d=-1$ . Set the new input to zero, then translate it back to the old input.
Step 3
Predict the old output
Use the point $(2,15)$ on the line. Starting at $t=15$ , a change of $d-2$ changes the predicted time by $-1(d-2)$ , so $t=15+(-1)(d-2)$ . Type $15+(-1)(d-2)$ below the regression; Desmos uses its fitted $d=-1$ and shows $t=18$ hours. Using $d=0$ instead would predict the old intercept, not the one asked for.
Step 4
Convert the prediction to the new output
Apply the new output rule $V=3(t-5)$ to the predicted $t=18$ . Type $3(18-5)$ ; Desmos shows $39$ . The same transformations carry the fitted line along with the data, so this is the $V$ value where $U=0$ : the new line's $V$ -intercept. Choice C.

Question 747·Hard·Math

Strategy

Hints

Step-by-step

Lessons that teach this

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Strategy

Hints

Step-by-step

Lessons that teach this