2 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 |
| ---- | ---- |
| 150 | 7 |
| 200 | 10 |
| 400 | 19 |
| 600 | 29 |
| 1000 | 51 |
write the linear regression equation for these data where miles driven is the independent variable. (round all values to the nearest hundredth.)

Question

Accepted Answer

# Explanation:
## Step1: Calculate the means
Let $x$ be miles driven and $y$ be number of gallons used.
$n = 5$
$\bar{x}=\frac{150 + 200+400+600+1000}{5}=\frac{2350}{5}=470$
$\bar{y}=\frac{7 + 10+19+29+51}{5}=\frac{116}{5}=23.2$
## Step2: Calculate the slope $m$
$$
\begin{align*}
m&=\frac{\sum_{i = 1}^{n}(x_{i}-\bar{x})(y_{i}-\bar{y})}{\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}}\
\sum_{i = 1}^{n}(x_{i}-\bar{x})(y_{i}-\bar{y})&=(150 - 470)(7- 23.2)+(200 - 470)(10 - 23.2)+(400 - 470)(19 - 23.2)+(600 - 470)(29 - 23.2)+(1000 - 470)(51 - 23.2)\
&=(-320)(-16.2)+(-270)(-13.2)+(-70)(-4.2)+(130)(5.8)+(530)(27.8)\
&=5184+3564 + 294+754+14734\
&=24530\
\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}&=(150 - 470)^{2}+(200 - 470)^{2}+(400 - 470)^{2}+(600 - 470)^{2}+(1000 - 470)^{2}\
&=(-320)^{2}+(-270)^{2}+(-70)^{2}+(130)^{2}+(530)^{2}\
&=102400+72900+4900+16900+280900\
&=478000
\end{align*}
$$
$m=\frac{24530}{478000}\approx0.05$
## Step3: Calculate the y - intercept $b$
$b=\bar{y}-m\bar{x}=23.2-0.05\times470=23.2 - 23.5=- 0.3$
# Answer:
$y = 0.05x-0.3$

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QUESTION IMAGE

Question

Explanation:

Step1: Calculate the means

Step2: Calculate the slope \(m\)

Step3: Calculate the y - intercept \(b\)

Answer: