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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.

A mathematics teacher wanted to see the correlation between test scores and homework. The homework grade (x) and test grade (y) are given in the accompanying table. Write the linear regression equation that represents this set of data, rounding all coefficients to the nearest tenth. Using this equation, find the projected test grade, to the nearest integer, for a student with a homework grade of 37.

Learn how to calculate a linear regression equation using the given homework and test grade data. Includes step-by-step calculations for slope and intercept, and predicts the test grade for a homework grade of 37. Suitable for students in Grades 9-12.

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