a linear model for the data is $y = 9.44t+84.182$. use the above scatter plot to decide whether the linear…

a linear model for the data is $y = 9.44t+84.182$. use the above scatter plot to decide whether the linear model fits the data well. the function is not a good model for the data the function is a good model for the data. estimate the retails sales in the u. s. in 2011. billions of dollars. use the model to predict the year that corresponds to retails sales of $242 billion. round down to the nearest year.

a linear model for the data is $y = 9.44t+84.182$. use the above scatter plot to decide whether the linear model fits the data well. the function is not a good model for the data the function is a good model for the data. estimate the retails sales in the u. s. in 2011. billions of dollars. use the model to predict the year that corresponds to retails sales of $242 billion. round down to the nearest year.

Answer

Explanation:

Step1: Check goodness - of - fit

Visually, the data points in the scatter - plot lie close to the line $y = 9.44t+84.182$. So, the function is a good model for the data.

Step2: Estimate 2011 retail sales

Assume $t = 0$ corresponds to some base year. If we assume $t$ is the number of years since a certain starting year, for 2011, we need to determine $t$. Let's assume the base - year is such that we can calculate $t$. But if we assume a simple start (e.g., $t = 0$ is 2000), then for 2011, $t = 11$. Substitute $t = 11$ into the equation $y=9.44t + 84.182$. $y=9.44\times11 + 84.182=9.44\times11+84.182=103.84+84.182 = 188.022$

Step3: Predict year for $y = 242$

Set $y = 242$ in the equation $y = 9.44t+84.182$. Then $242=9.44t + 84.182$. First, subtract 84.182 from both sides: $242−84.182=9.44t$, so $157.818 = 9.44t$. Then solve for $t$: $t=\frac{157.818}{9.44}\approx16.72$. Rounding down, $t = 16$.

Answer:

The function is a good model for the data. 188.022 16