Eric is preparing for a regional swim meet that is a month away. He swims the same distance and clocks his best time on several different days. He plots the best time (in minutes) he swam each day as his y-variable and the day he swam it as his x-variable. When he plots his times and finds a line of best fit, he gets the equation y=−0.03x+3.45.

If Eric continues to practice, on the 15th day his best time for the day should be _[blank]_ minutes.

Do not round your answer.

On the 15th day Eric's best time for the day should be 3.00 minutes.

Step-by-step explanation:

The line of best also known as the least square regression line is used to predict the value of the dependent variable based on a single independent variable.

The general form of a line of best is:

y = b X + a

Here,

y = dependent variable

x = independent variable

a = intercept

b = slope

In this case the dependent variable is defined as the best time (in minutes) Eric swam each day and the dependent variable is defined as the nth day.

The line of best fit is:

y = -0.03 x + 3.45

Compute the value of y for x = 15 as follows:

y = -0.03 x + 3.45

  = (-0.03 × 15) + 3.45

  = -0.45 + 3.45

  = 3.00

Thus, on the 15th day Eric's best time for the day should be 3.00 minutes.