QUESTION IMAGE
Question
emma recently purchased a new car. she decided to keep track of how many gallons of gas she used on five of her business trips. the results are shown in the table below.
| miles driven | number of gallons used |
|---|---|
| 200 | 10 |
| 400 | 19 |
| 600 | 29 |
| 1000 | 51 |
round your answer to the nearest hundredth. what is the y-intercept of this equation?
Step1: Define Variables
Let \( x \) be the miles driven (independent variable) and \( y \) be the number of gallons used (dependent variable). We assume a linear relationship \( y = mx + b \), where \( m \) is the slope and \( b \) is the y - intercept.
Step2: Calculate the Slope \( m \)
We can use two points to calculate the slope. Let's take the first two points \((x_1,y_1)=(150,7)\) and \((x_2,y_2)=(200,10)\). The formula for the slope is \( m=\frac{y_2 - y_1}{x_2 - x_1}\)
\( m=\frac{10 - 7}{200 - 150}=\frac{3}{50}=0.06\)
We can also use the formula for the slope of a linear regression (since we have multiple points) \( m=\frac{n\sum_{i = 1}^{n}x_iy_i-\sum_{i = 1}^{n}x_i\sum_{i = 1}^{n}y_i}{n\sum_{i = 1}^{n}x_i^2-(\sum_{i = 1}^{n}x_i)^2}\)
First, calculate the necessary sums:
- \( n = 5 \)
- \( \sum_{i=1}^{5}x_i=150 + 200+400 + 600+1000=2350 \)
- \( \sum_{i=1}^{5}y_i=7 + 10+19 + 29+51=116 \)
- \( \sum_{i=1}^{5}x_iy_i=(150\times7)+(200\times10)+(400\times19)+(600\times29)+(1000\times51)=1050+2000 + 7600+17400+51000=79050 \)
- \( \sum_{i=1}^{5}x_i^2=(150^2)+(200^2)+(400^2)+(600^2)+(1000^2)=22500+40000 + 160000+360000+1000000=1582500 \)
Now, substitute into the slope formula:
\( m=\frac{5\times79050-2350\times116}{5\times1582500-(2350)^2}\)
\(=\frac{395250 - 272600}{7912500-5522500}\)
\(=\frac{122650}{2390000}\approx0.0513\) (using linear regression gives a more accurate slope)
Step3: Calculate the y - intercept \( b \)
We use the formula \( b=\bar{y}-m\bar{x}\), where \( \bar{x}=\frac{\sum_{i = 1}^{n}x_i}{n}\) and \( \bar{y}=\frac{\sum_{i = 1}^{n}y_i}{n}\)
\( \bar{x}=\frac{2350}{5} = 470\)
\( \bar{y}=\frac{116}{5}=23.2\)
Using the linear regression slope \( m\approx0.0513\)
\( b=23.2-0.0513\times470\)
\(=23.2 - 24.111\)
\(=- 0.911\approx - 0.91\)
If we use the slope from the first two points \( m = 0.06\)
\( b=7-0.06\times150=7 - 9=- 2\) (but linear regression is more accurate for multiple points)
Using the linear regression method (more appropriate for 5 points):
We can also use the equation of the line \( y=mx + b\), and solve for \( b\) using one of the points. Let's use the point \((150,7)\) and \( m\approx0.0513\)
\( 7=0.0513\times150 + b\)
\( 7 = 7.695+b\)
\( b=7 - 7.695=- 0.695\approx - 0.70\) (Wait, there was a miscalculation in the linear regression slope earlier)
Let's recalculate the slope correctly:
\( n = 5 \)
\( \sum x=150 + 200+400+600+1000 = 2350\)
\( \sum y=7 + 10+19+29+51=116\)
\( \sum xy=150\times7 + 200\times10+400\times19+600\times29+1000\times51=1050 + 2000+7600+17400+51000 = 79050\)
\( \sum x^{2}=150^{2}+200^{2}+400^{2}+600^{2}+1000^{2}=22500 + 40000+160000+360000+1000000 = 1582500\)
\( m=\frac{n\sum xy-\sum x\sum y}{n\sum x^{2}-(\sum x)^{2}}=\frac{5\times79050-2350\times116}{5\times1582500 - 2350^{2}}\)
\( 5\times79050 = 395250\)
\( 2350\times116=2350\times(100 + 16)=235000+37600 = 272600\)
\( 5\times1582500=7912500\)
\( 2350^{2}=(2000 + 350)^{2}=2000^{2}+2\times2000\times350+350^{2}=4000000+1400000 + 122500=5522500\)
\( m=\frac{395250-272600}{7912500 - 5522500}=\frac{122650}{2390000}\approx0.0513\) (this is correct)
\( \bar{x}=\frac{2350}{5}=470\), \( \bar{y}=\frac{116}{5} = 23.2\)
\( b=\bar{y}-m\bar{x}=23.2-0.0513\times470\)
\( 0.0513\times470 = 0.0513\times(400+70)=0.0513\times400+0.0513\times70 = 20.52+3.591 = 24.111\)
\( b=23.2 - 24.111=- 0.911\approx - 0.91\)
But if we use a calculator for linear regression (using all 5 points):
The linear regression equation is of the form \( y = ax + b\), where \( a\) is the slope and \( b\) is the y - intercept.
Using a calculat…
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\(-0.91\)