Question

alex used a regression calculator to find the equation of the trend line ( hat{y} approx ax + b ) for the data in the table. he rounded the values for ( a ) and ( b ) on his calculator as shown below.( a approx 1.7 )( b approx 6.2 )which trend line compares the data in the table?( \bigcirc s approx 1.7f + 6.2 )( \bigcirc s approx 6.2f + 1.7 )( \bigcirc f approx 1.7s + 6.2 )( \bigcirc f approx 6.2s + 1.7 )

Explanation:

Step1: Recall the trend line formula

The general form of the trend line is \( y = ax + b \), where \( a \approx 1.7 \) and \( b \approx 6.2 \). We need to identify which variable is the independent (\( x \)) and which is the dependent (\( y \)) from the table. Looking at the table, the first column (let's say \( t \) or \( f \)) and the second column (let's say \( s \)) – the trend line should model the relationship between them. The given \( a = 1.7 \) and \( b = 6.2 \), so the equation should be \( s = 1.7f + 6.2 \) (since the first option matches the form \( y = ax + b \) with \( y = s \), \( x = f \), \( a = 1.7 \), \( b = 6.2 \)).

Step2: Eliminate other options

• Option 2: \( s = 6.2f + 1.7 \) has \( a = 6.2 \) and \( b = 1.7 \), which doesn't match the given \( a \) and \( b \).
• Option 3: \( f = 1.7s + 6.2 \) swaps the variables and has incorrect coefficients.
• Option 4: \( f = 0.2s + 1.7 \) has wrong coefficients and variable swap.

Answer:

A. \( s \approx 1.7f + 6.2 \)