1990 55.8
1994 47.2
1996 40.1
1999 42.4
2001 47.5
a. Code the years using t = 0 for 1990 and fit a quadratic function to the data with the regression program on your calculator. Equation is.
b. Predict the maintenance and repair costs in 2010.
I don't have a graphing calculator, can someone help me with this how to do this...
type in graphing calculator in a search engine and it brings up quite a few online free ones
yeah but i can't figure out how to use those...
To fit a quadratic function to the given data and find the equation, follow these steps:
1. Write down the data points in two separate columns: one for the years and the other for the corresponding maintenance and repair costs.
2. Assign a value of 0 (t = 0) to the year 1990 and increment the values by 1 for each subsequent year. This will create a new column of values for t.
3. Square the values in the t column to obtain a new column of t^2 values.
4. Create a new column by multiplying the t values with the maintenance and repair cost values. This will result in a column of t * maintenance and repair costs.
5. Sum up all the values in the t, t^2, and t * maintenance and repair cost columns.
6. Use these sums to find the coefficients of the quadratic equation. The equation will be in the form: y = at^2 + bt + c.
The coefficient a can be calculated using the formula:
a = (C * Σt^2 - Σt * Σt * maintenance and repair costs) / (C * Σt^2 - (Σt)^2)
The coefficient b can be calculated using the formula:
b = (C * Σt * maintenance and repair costs - Σt * Σt * maintenance and repair costs) / (C * Σt^2 - (Σt)^2)
The coefficient c can be calculated using the formula:
c = (Σt * Σt * maintenance and repair costs * Σt^2 - Σt * maintenance and repair costs * Σt * Σt * maintenance and repair costs) / (C * Σt^2 - (Σt)^2)
Where C is the number of data points.
Now let's calculate the coefficients and find the equation.
Step 1: Data Points
1990 55.8
1994 47.2
1996 40.1
1999 42.4
2001 47.5
Step 2: Assigning values of t
1990 -> t = 0
1994 -> t = 4
1996 -> t = 6
1999 -> t = 9
2001 -> t = 11
Step 3: Squaring t
t^2 = 0^2, 4^2, 6^2, 9^2, 11^2 = 0, 16, 36, 81, 121
Step 4: Calculating t * maintenance and repair costs
t * maintenance and repair costs = 0 * 55.8, 4 * 47.2, 6 * 40.1, 9 * 42.4, 11 * 47.5 = 0, 188.8, 240.6, 381.6, 522.5
Step 5: Summing up the values
Σt = 0 + 4 + 6 + 9 + 11 = 30
Σt^2 = 0 + 16 + 36 + 81 + 121 = 254
Σ(t * maintenance and repair costs) = 0 + 188.8 + 240.6 + 381.6 + 522.5 = 1333.5
Step 6: Calculating the coefficients
C = 5 (number of data points)
a = (C * Σt^2 - Σt * Σt * maintenance and repair costs) / (C * Σt^2 - (Σt)^2)
= (5 * 254 - 30 * 1333.5) / (5 * 254 - 900)
= -0.01587
b = (C * Σt * maintenance and repair costs - Σt * Σt * maintenance and repair costs) / (C * Σt^2 - (Σt)^2)
= (5 * 1333.5 - 30 * 1333.5) / (5 * 254 - 900)
= -4.581
c = (Σt * Σt * maintenance and repair costs * Σt^2 - Σt * maintenance and repair costs * Σt * Σt * maintenance and repair costs) / (C * Σt^2 - (Σt)^2)
= (30 * 1333.5 * 254 - 30 * 1333.5 * 30 * 1333.5) / (5 * 254 - 900)
= 1179.286
Therefore, the equation for the quadratic function is:
y = -0.01587t^2 - 4.581t + 1179.286
To predict the maintenance and repair costs in 2010, we need to find the value of t for the year 2010. Since t = 0 corresponds to the year 1990, we can calculate the value of t for the year 2010 as follows:
2010 - 1990 = 20
Now substitute this value of t into the equation we found earlier:
y = -0.01587(20)^2 - 4.581(20) + 1179.286
= -6.348 - 91.62 + 1179.286
= 1081.318
The predicted maintenance and repair costs in 2010 would be approximately 1081.318.