The annual profits for a company are given in the following table, where x represents the number of years since 2004, 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 hundredth*. Using this equation, estimate the *calendar* year in which the profits would reach 304 thousand dollars.

Years since 2004 (x)Profits (y)
(in thousands of dollars)

0 151

1 147

2 200

3 223

To find the linear regression equation that represents this set of data, we first need to determine the line of best fit for the points. The linear regression equation has the form:

\[
y = mx + b
\]

where:
- \( y \) is the profit in thousands of dollars.
- \( x \) is the number of years since 2004.
- \( m \) is the slope of the line.
- \( b \) is the y-intercept.

### Step 1: Calculate the Slope (m) and Y-Intercept (b)

The formulas for the slope \( m \) and y-intercept \( b \) are given by:

\[
m = \frac{n(\sum xy) - (\sum x)(\sum y)}{n(\sum x^2) - (\sum x)^2}
\]

\[
b = \frac{(\sum y)(\sum x^2) - (\sum x)(\sum xy)}{n(\sum x^2) - (\sum x)^2}
\]

where:
- \( n \) is the number of data points.
- \( \sum x \) is the sum of all x-values.
- \( \sum y \) is the sum of all y-values.
- \( \sum xy \) is the sum of the product of each pair of x and y values.
- \( \sum x^2 \) is the sum of the squares of each x-value.

Let's calculate these values:

\[
\text{Data points: } (0, 151), (1, 147), (2, 200), (3, 223)
\]

Now we compute the necessary sums:

\[
\sum x = 0 + 1 + 2 + 3, \quad \sum y = 151 + 147 + 200 + 223, \quad \sum xy = (0)(151) + (1)(147) + (2)(200) + (3)(223)
\]
\[
\sum x^2 = (0)^2 + (1)^2 + (2)^2 + (3)^2
\]

Let's plug in the values and calculate the slope \( m \) and intercept \( b \).The linear regression equation that represents the given set of data, with coefficients rounded to the nearest hundredth, is:

\[
y = 26.9x + 139.9
\]

### Step 2: Estimate the Year When Profits Reach 304 Thousand Dollars

We need to find the value of \( x \) (years since 2004) when the profit \( y \) is 304 thousand dollars. Substituting \( y = 304 \) into the equation:

\[
304 = 26.9x + 139.9
\]

Let's solve for \( x \).The estimated value of \( x \) when the profits reach 304 thousand dollars is approximately \( 6.10 \).

### Step 3: Convert to Calendar Year

Since \( x \) represents the number of years since 2004, the calendar year can be found by adding \( x \) to 2004:

\[
\text{Year} = 2004 + 6.10 \approx 2010.1
\]

Thus, the profits are estimated to reach 304 thousand dollars in early 2010.

Would you like more details or have any questions?

### Related Questions

1. How is the slope of a line related to the change in profit over time?
2. How can the linear regression model be used to predict future values?
3. What other types of regression models might fit this data better?
4. What is the significance of the y-intercept in this context?
5. How would the prediction change if additional data points were included?

**Tip:** Linear regression is most effective when the data shows a roughly linear trend; otherwise, consider using other forms of regression or curve fitting!

The annual profits for a company are given in the following table, where x represents the number of years since 2004, 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 hundredth. Using this equation, estimate the calendar year in which the profits would reach 304 thousand dollars.

This problem involves finding the linear regression equation to model a company's annual profits from 2004, given data for years since 2004. Learn how to calculate the slope and intercept of the line and estimate the year when the profits will reach 304 thousand dollars, using algebra and statistics concepts. Suitable for Grades 10-12.

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