An employer uses the linear regression equation y = 0.18 x + 320.22 to predict the weekly salary, y, of an employee who sells x dollars worth of merchandise. Last week, Joaquin sold $1500 worth of merchandise. The same week, Alex earned $650. Using the regression equation, which is an accurate comparison?

Question

Alex sold merchandise worth about $60 more than what Joaquin sold.
Joaquin sold merchandise worth about $60 more than what Alex sold.
Alex earned about $60 more than Joaquin did.
Joaquin earned about $60 more than Alex did.

Answer 1

To compare Joaquin's and Alex's earnings using the linear regression equation \( y = 0.18x + 320.22 \), we first need to calculate Joaquin's predicted salary based on the amount of merchandise he sold, which is $1500.

Plugging in the values into the regression equation:

\[ y = 0.18(1500) + 320.22 \]

Calculating \( 0.18 \times 1500 \):

\[ 0.18 \times 1500 = 270 \]

Now, substituting this back into the equation:

\[ y = 270 + 320.22 = 590.22 \]

So, Joaquin's predicted weekly salary is approximately $590.22.

Now we can compare this with Alex's earnings, which are given as $650.

Next, we find the difference in earnings:

\[ 650 - 590.22 \approx 59.78 \]

This result indicates that Alex earned about $60 more than Joaquin.

Thus, the accurate comparison from the options given is:

Alex earned about $60 more than Joaquin did.