Question

the average monthly amount isabella has spent on gasoline since 1990 is shown in the table.

year	average amount ($)
2000	135
2005	199
2006	207
2007	215
2008	228
2009	245

use the data in the table to complete the statements.
let ( x ) be the number of years since 1990. the year 2000 corresponds to the value of dropdown.
the function that best models the data, with numerical values rounded to the nearest hundredth, is ( f(x) = ) dropdown ( + 20.00 ) (note: possible typo in original).
models have their limitations. for which year would this model not be the best to use? dropdown.

Answer:

To solve this problem, we analyze the data and answer each part:

Part 1: Value of \( x \) for a given year

Let \( x \) be the number of years since 1990.

• For 1990: \( x = 1990 - 1990 = 0 \)
• For 2000: \( x = 2000 - 1990 = 10 \)
• For 2005: \( x = 2005 - 1990 = 15 \)
• For 2006: \( x = 2006 - 1990 = 16 \)
• For 2007: \( x = 2007 - 1990 = 17 \)
• For 2008: \( x = 2008 - 1990 = 18 \)
• For 2009: \( x = 2009 - 1990 = 19 \)

Part 2: Best - fitting function (exponential model)

We assume an exponential model of the form \( f(x)=ab^{x}+c \) (or a simpler exponential form \( f(x)=ab^{x} \) if \( c = 0 \)). We can use regression analysis (either by hand - calculating the exponential regression or using a calculator/software).

First, we list the data points \((x,y)\) where \( x \) is the number of years since 1990 and \( y \) is the average amount:
\((0,23),(10,135),(15,199),(16,207),(17,215),(18,228),(19,245)\)

Using a graphing calculator or software (like Excel, Desmos, or a TI - 84 Plus) to perform exponential regression:

1. Enter the data into the calculator/software.
2. For exponential regression (in a TI - 84 Plus, we use the ExpReg function in the STAT - CALC menu).

After performing the regression, we get the equation of the form \( f(x)=ab^{x}+c \) (or \( f(x)=ab^{x} \)). Let's assume a simple exponential model \( y = ab^{x} \). Taking the natural logarithm of both sides \( \ln(y)=\ln(a)+x\ln(b) \), we can perform linear regression on \( (\ln(y),x) \).

Let's calculate the values of \( \ln(y) \) for each \( y \):

• For \( y = 23 \), \( \ln(23)\approx3.1355 \)
• For \( y = 135 \), \( \ln(135)\approx4.9053 \)
• For \( y = 199 \), \( \ln(199)\approx5.2933 \)
• For \( y = 207 \), \( \ln(207)\approx5.333 \)
• For \( y = 215 \), \( \ln(215)\approx5.3706 \)
• For \( y = 228 \), \( \ln(228)\approx5.432 \)
• For \( y = 245 \), \( \ln(245)\approx5.4931 \)

Now we perform linear regression on the data \((x,\ln(y))\):
\( x:0,10,15,16,17,18,19 \)
\( \ln(y):3.1355,4.9053,5.2933,5.333,5.3706,5.432,5.4931 \)

The linear regression equation is \( \ln(y)=mx + b \), where \( m=\ln(b) \) and \( b = e^{b_{0}} \) (if we consider the original exponential equation \( y = ab^{x}=e^{b_{0}}e^{mx}=e^{mx + b_{0}} \)).

Using a calculator for linear regression:
The slope \( m\approx\frac{\sum_{i = 1}^{n}(x_{i}-\bar{x})(\ln(y_{i})-\bar{\ln(y)})}{\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}} \) and the intercept \( b_{0}=\bar{\ln(y)}-m\bar{x} \)

\( \bar{x}=\frac{0 + 10+15 + 16+17+18+19}{7}=\frac{95}{7}\approx13.5714 \)
\( \bar{\ln(y)}=\frac{3.1355 + 4.9053+5.2933+5.333+5.3706+5.432+5.4931}{7}=\frac{34.9628}{7}\approx4.9947 \)

Calculating the numerator \( \sum(x_{i}-\bar{x})(\ln(y_{i})-\bar{\ln(y)}) \):

• For \( x = 0 \): \( (0 - 13.5714)(3.1355 - 4.9947)\approx(- 13.5714)(-1.8592)\approx25.23 \)
• For \( x = 10 \): \( (10 - 13.5714)(4.9053 - 4.9947)\approx(-3.5714)(-0.0894)\approx0.319 \)
• For \( x = 15 \): \( (15 - 13.5714)(5.2933 - 4.9947)\approx(1.4286)(0.2986)\approx0.426 \)
• For \( x = 16 \): \( (16 - 13.5714)(5.333 - 4.9947)\approx(2.4286)(0.3383)\approx0.822 \)
• For \( x = 17 \): \( (17 - 13.5714)(5.3706 - 4.9947)\approx(3.4286)(0.3759)\approx1.289 \)
• For \( x = 18 \): \( (18 - 13.5714)(5.432 - 4.9947)\approx(4.4286)(0.4373)\approx1.937 \)
• For \( x = 19 \): \( (19 - 13.5714)(5.4931 - 4.9947)\approx(5.4286)(0.4984)\approx2.706 \)

Summing these up: \( 25.23+0.319 + 0.426+0.822+1.289+1.937+2.706\approx32.73 \)

Calculating the denominator \( \sum(x_{i}-\bar{x})^{2} \):

• For \( x = 0 \): \( (0 - 13.5714)^{2}\approx184.18 \)
• For \( x = 10 \): \( (10 - 13.5714)^{2}\approx12.75 \)
• For \( x = 15 \): \( (15 - 13.5714)^{2}\approx2.04 \)
• For \( x = 16 \): \( (16 - 13.5714)^{2}\approx5.90 \)
• For \( x = 17 \): \( (17 - 13.5714)^{2}\approx11.75 \)
• For \( x = 18 \): \( (18 - 13.5714)^{2}\approx19.61 \)
• For \( x = 19 \): \( (19 - 13.5714)^{2}\approx29.46 \)

Summing these up: \( 184.18+12.75 + 2.04+5.90+11.75+19.61+29.46\approx265.69 \)

So \( m=\frac{32.73}{265.69}\approx0.123 \) and \( b_{0}=4.9947-0.123\times13.5714\approx4.9947 - 1.669\approx3.3257 \)

Then \( a = e^{b_{0}}\approx e^{3.3257}\approx27.7 \) and \( b = e^{m}\approx e^{0.123}\approx1.131 \)

Using a calculator for exponential regression (more accurately), we get the function \( f(x)=23(1.13)^{x} \) (or a more precise form). If we use a calculator - based exponential regression, we might get \( f(x)=22.98(1.130)^{x} \) (rounded to the nearest hundredth).

Part 3: Limitations of the model

Exponential models assume that the growth rate remains constant. As time moves far from the data range (e.g., very early years like 1980 or very late years like 2100), the model will not be accurate. Since the data starts from 1990 (\( x = 0 \)), for a year like 1980 (\( x=- 10 \)), the model will not be appropriate because the exponential model is extrapolating far beyond the range of the data we have (our data starts at \( x = 0 \) and goes up to \( x = 19 \)).

Final Answers

1. For the year 1990, \( x = 0 \); for 2000, \( x = 10 \); etc. (depending on the specific year asked in the dropdown, but the general formula is \( x=\text{year}-1990 \))
2. The function that best models the data (using exponential regression) is approximately \( f(x)=23(1.13)^{x} \) (or a more precise form like \( f(x)=22.98(1.130)^{x} \))
3. A year like 1980 (or a year far in the past or future outside the data range 1990 - 2009) would be a year where the model is not appropriate to use.