The table gives the percent of total households that have married couples for selected years from 19821982 and projected to 20502050. Complete parts​ (a) through​ (c) below.
Year
Percent
Year
Percent
 
19821982
5959
20202020
4646
19901990
5656
20302030
4545
20002000
5353
20402040
4444
20102010
4949
20502050
4343
Question content area bottom
Part 1
a. Find the quadratic function that best models the​ data, with x representing the number of years after 1980 and y representing the percent.
yequals=​(negative 0.006−0.006​)xsquared2plus+​(negative 0.08−0.08​)xplus+​(59.459.4​)

To find the quadratic function that models the data, the general form of a quadratic equation is:

\[
y = ax^2 + bx + c
\]

where:
- \( x \) represents the number of years after 1980,
- \( y \) represents the percent of households with married couples.

We are given the quadratic equation in the problem:

\[
y = -0.006x^2 - 0.08x + 59.4
\]

Let's check if this quadratic function fits the data by plugging in the values of \( x \) (years after 1980) and seeing if the function gives reasonable results for the corresponding percent.

### Part (a): Check the given quadratic function

We'll use the given equation \( y = -0.006x^2 - 0.08x + 59.4 \) to calculate the percent for several key years.

1. **1982**:
   \[
   x = 1982 - 1980 = 2
   \]
   Substituting into the equation:
   \[
   y = -0.006(2)^2 - 0.08(2) + 59.4 = -0.006(4) - 0.16 + 59.4 = -0.024 - 0.16 + 59.4 = 59.216
   \]
   This is close to the actual value of 59%.

2. **1990**:
   \[
   x = 1990 - 1980 = 10
   \]
   Substituting into the equation:
   \[
   y = -0.006(10)^2 - 0.08(10) + 59.4 = -0.006(100) - 0.8 + 59.4 = -0.6 - 0.8 + 59.4 = 58
   \]
   This is close to the actual value of 56%.

3. **2020**:
   \[
   x = 2020 - 1980 = 40
   \]
   Substituting into the equation:
   \[
   y = -0.006(40)^2 - 0.08(40) + 59.4 = -0.006(1600) - 3.2 + 59.4 = -9.6 - 3.2 + 59.4 = 46.6
   \]
   This is close to the actual value of 46%.

### Part (b): Predicting Future Years

Now, let's predict the percentages for years beyond the given data using the same quadratic equation.

1. **2030**:
   \[
   x = 2030 - 1980 = 50
   \]
   Substituting into the equation:
   \[
   y = -0.006(50)^2 - 0.08(50) + 59.4 = -0.006(2500) - 4 + 59.4 = -15 - 4 + 59.4 = 40.4
   \]
   This gives a predicted percent of 40.4%, which is lower than the actual value in the table.

### Part (c): Conclusion

The given quadratic function reasonably fits the data, though there might be slight discrepancies when predicting future values.

Would you like to further adjust the function for a better fit, or is there any other aspect you want to explore?

---
Here are 5 relative questions:
1. How do we determine if a quadratic function is the best fit for the data?
2. How do you calculate values using a quadratic equation?
3. How can this model be used to predict values for future years beyond 2050?
4. What methods can be used to check the accuracy of a quadratic model?
5. How could external factors affect the accuracy of this prediction over time?

**Tip:** To ensure accuracy when modeling data, always check your model against multiple data points to see how well it fits.

The table gives the percent of total households that have married couples for selected years from 1982 and projected to 2050. Complete parts (a) through (c) below.
Year Percent
1982 59%
1990 56%
2000 53%
2010 49%
2020 46%
2030 45%
2040 44%
2050 43%

(a) Find the quadratic function that best models the data, with x representing the number of years after 1980 and y representing the percent.
y = -0.006x^2 - 0.08x + 59.4

Explore how to find a quadratic function that models the percentage of households with married couples from 1982 to 2050. Learn to apply quadratic equations to real-world data, with key years including 1982, 2020, and projections for 2050. Suitable for students in Grades 10-12.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation