To solve this problem, we need to perform a linear regression on the given data points, where $x$ represents the homework grade and $y$ represents the test grade. Here's a step-by-step guide:

1. Table of data:

   • Homework Grade ($x$): 71, 87, 85, 73, 73, 90, 86
   • Test Grade ($y$): 65, 86, 80, 78, 68, 77, 90

2. Linear Regression Equation: The linear regression equation takes the form: $y = mx + b$ where:

   • $m$ is the slope of the line.
   • $b$ is the y-intercept.

3. Formulas for slope $m$ and intercept $b$: The slope $m$ and y-intercept $b$ can be calculated as: $m = \frac{n \sum xy - \sum x \sum y}{n \sum x^2 - (\sum x)^2}$ $b = \frac{\sum y - m \sum x}{n}$ where:

   • $n$ is the number of data points.
   • $\sum xy$ is the sum of the products of $x$ and $y$.
   • $\sum x$ is the sum of the homework grades.
   • $\sum y$ is the sum of the test grades.
   • $\sum x^2$ is the sum of the squares of the homework grades.

Let me calculate this step-by-step to find the linear regression equation.

Step 1: Calculating necessary sums

Given:

• $n = 7$ (7 data points),

$\sum x = 71 + 87 + 85 + 73 + 73 + 90 + 86 = 565$ $\sum y = 65 + 86 + 80 + 78 + 68 + 77 + 90 = 544$ $\sum x^2 = 71^2 + 87^2 + 85^2 + 73^2 + 73^2 + 90^2 + 86^2 = 50419$ $\sum xy = (71 \times 65) + (87 \times 86) + (85 \times 80) + (73 \times 78) + (73 \times 68) + (90 \times 77) + (86 \times 90) = 44545$

Step 2: Calculating slope $m$ and intercept $b$

Using the formulas: $m = \frac{n \sum xy - \sum x \sum y}{n \sum x^2 - (\sum x)^2}$ $m = \frac{7(44545) - (565)(544)}{7(50419) - (565)^2}$ $m = \frac{311815 - 307760}{352933 - 319225}$ $m = \frac{4055}{33708} \approx 0.12$

Now, for the intercept: $b = \frac{\sum y - m \sum x}{n}$ $b = \frac{544 - (0.12)(565)}{7}$ $b = \frac{544 - 67.8}{7} = \frac{476.2}{7} \approx 68.0$

Step 3: Equation of the line

Thus, the linear regression equation is approximately: $y = 0.12x + 68.0$

Step 4: Predicting the test grade for a homework grade of 37

Substitute $x = 37$ into the equation: $y = 0.12(37) + 68.0 = 4.44 + 68.0 = 72.44$ Rounding to the nearest integer, the predicted test grade is: $\boxed{72}$

Would you like more details on any part of this process?

Here are 5 related questions to explore:

1. How does linear regression help in predicting future values?
2. How would the prediction change if the homework grade were higher, say 95?
3. How do changes in the slope $m$ affect the accuracy of the prediction?
4. Can this linear model be used for all homework-test relationships, or are there limits?
5. How does the choice of rounding in coefficients impact the predicted results?

Tip: When using linear regression, it's important to check the correlation strength between variables to ensure the model is reliable for predictions.