The annual profits for a company are given in the following table, where x represents the number of years since 2011, and y represents the profit in thousands of dollars. Write the linear regression equation that represents this set of data, rounding all coefficients to the nearest tenth. Using this equation, estimate the calendar year in which the profits would reach 248 thousand dollars.
Years since 2011 (x)	Profits (y)
(in thousands of dollars)
00	9595
11	133133
22	131131
33	163163
Answer
Attempt 2 out of 2

Regression Equation: 

Final Answer: 


Let's go through the steps to find the linear regression equation for the given data.

### Step 1: Represent the Data
The data is presented as:
- $ x = 0, y = 95 $
- $ x = 1, y = 133 $
- $ x = 2, y = 131 $
- $ x = 3, y = 163 $

where $ x $ is the number of years since 2011, and $ y $ is the profit in thousands of dollars.

### Step 2: Calculating the Linear Regression Equation
The formula for a linear regression equation is:
\[
y = mx + b
\]
where:
- $ m $ is the slope of the line.
- $ b $ is the y-intercept.

To find the slope ($ m $), we use the formula:
\[
m = \frac{n \sum xy - (\sum x)(\sum y)}{n \sum x^2 - (\sum x)^2}
\]
and to find the intercept ($ b $):
\[
b = \frac{\sum y - m \sum x}{n}
\]
where $ n $ is the number of data points.

Let's calculate each term:

- $ n = 4 $ (since there are 4 data points)
- $ \sum x = 0 + 1 + 2 + 3 = 6 $
- $ \sum y = 95 + 133 + 131 + 163 = 522 $
- $ \sum xy = (0)(95) + (1)(133) + (2)(131) + (3)(163) = 0 + 133 + 262 + 489 = 884 $
- $ \sum x^2 = 0^2 + 1^2 + 2^2 + 3^2 = 0 + 1 + 4 + 9 = 14 $

Substituting these values into the slope formula:
\[
m = \frac{4(884) - (6)(522)}{4(14) - (6)^2} = \frac{3536 - 3132}{56 - 36} = \frac{404}{20} = 20.2
\]

Now calculate the intercept ($ b $):
\[
b = \frac{522 - 20.2(6)}{4} = \frac{522 - 121.2}{4} = \frac{400.8}{4} = 100.2
\]

### Step 3: Write the Linear Regression Equation
The linear regression equation is:
\[
y = 20.2x + 100.2
\]

### Step 4: Estimate the Year When Profits Reach 248 Thousand Dollars
We need to find $ x $ when $ y = 248 $:
\[
248 = 20.2x + 100.2
\]
Solving for $ x $:
\[
20.2x = 248 - 100.2
\]
\[
20.2x = 147.8
\]
\[
x = \frac{147.8}{20.2} \approx 7.3
\]

Since $ x $ represents the number of years since 2011:
\[
\text{Estimated year} = 2011 + 7.3 \approx 2018
\]

### Final Answers:
- **Regression Equation**: $ y = 20.2x + 100.2 $
- **Estimated Year**: 2018

Would you like further details or have any questions about these steps?

Here are five related questions to consider:
1. How do you interpret the slope of the regression line in this context?
2. What is the significance of the y-intercept in the regression equation?
3. How would the prediction change if the profit was expected to be 300 thousand dollars?
4. What are some limitations of using linear regression for predicting future values?
5. How could we measure the accuracy of this regression model?

**Tip:** When rounding coefficients in regression analysis, always consider the context and the degree of precision required for practical applications.

The annual profits for a company are given in the following table, where x represents the number of years since 2011, and y represents the profit in thousands of dollars. Write the linear regression equation that represents this set of data, rounding all coefficients to the nearest tenth. Using this equation, estimate the calendar year in which the profits would reach 248 thousand dollars.
Years since 2011 (x)	Profits (y)
(in thousands of dollars)
00	95
11	133
22	131
33	163

Learn how to create a linear regression equation to predict the future profits of a company based on past data. This example shows how to estimate when the company's profits will reach $248,000. Suitable for high school students in Grades 10-12.

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