Question

1. the data in the table below shows the average temperature in northern latitudes:

latitude (°n) 0 10 20 30 40 50 60 70 80
temp (°f) 79 81 79 68 58 43 28 13 1

a) find the line of best fit:

b) estimate the average temperature for a city with a latitude of 48°:

1. the data in the table below shows the number of passengers and number of suitcases on various airplanes.

passengers 75 92 115 128 143 154 178 200
suitcases 159 180 239 272 290 310 357 405

a) find the line of best fit:

b) estimate the number of suitcases on a flight carrying 250 people.

1. the data in the table below shows the number of graduating seniors at canyon valley high school since 2012.

year 2012 2013 2014 2015 2016 2017
graduates 340 348 356 361 375 387

a) find the line of best fit:

b) estimate the number of graduating seniors in 2025.

1. the data in the table to the left shows the olympic 500 - meter gold medal speed skating times.

year time (s)
1980 42.2
1984 43.2
1988 40.4
1992 42.0
1994 39.5
1998 38.2

a) find the line of best fit:

b) estimate the 500 - meter time for the 2020 olympics.

1. the data in the table to the left shows sales for a certain department store (in billions of dollars).

a) find the line of best fit
b) estimate the sales total in 2025.

Explanation:

1.

Step1: Calculate sums for formula

Let $x$ be latitude and $y$ be temperature. Calculate $\sum x$, $\sum y$, $\sum x^2$, $\sum xy$, $n$ (number of data - points, $n = 9$).
$\sum x=0 + 10+20+\cdots+80 = 360$
$\sum y=79 + 81+79+\cdots+1 = 350$
$\sum x^2=0^2+10^2+20^2+\cdots+80^2 = 20400$
$\sum xy=0\times79 + 10\times81+20\times79+\cdots+80\times1 = 1120$

Step2: Calculate slope $m$

The formula for the slope $m$ of the line of best - fit $y=mx + b$ is $m=\frac{n\sum xy-\sum x\sum y}{n\sum x^2-(\sum x)^2}$
$m=\frac{9\times1120 - 360\times350}{9\times20400-360^2}=\frac{10080 - 126000}{183600 - 129600}=\frac{- 115920}{54000}=- 2.1467$

Step3: Calculate y - intercept $b$

The formula for the y - intercept $b$ is $b=\frac{\sum y-m\sum x}{n}$
$b=\frac{350-(-2.1467)\times360}{9}=\frac{350 + 772.812}{9}=\frac{1122.812}{9}=124.757$
The line of best - fit is $y=-2.1467x + 124.757$

Step4: Estimate temperature for $x = 48$

Substitute $x = 48$ into $y=-2.1467x + 124.757$
$y=-2.1467\times48+124.757=-103.0416 + 124.757 = 21.7154\approx21.72$

Step1: Calculate sums

Let $x$ be the number of passengers and $y$ be the number of suitcases. $n = 8$
$\sum x=75 + 92+115+\cdots+200 = 1085$
$\sum y=159 + 180+239+\cdots+405 = 2212$
$\sum x^2=75^2+92^2+115^2+\cdots+200^2 = 167379$
$\sum xy=75\times159+92\times180+115\times239+\cdots+200\times405 = 337935$

Step2: Calculate slope $m$

$m=\frac{n\sum xy-\sum x\sum y}{n\sum x^2-(\sum x)^2}=\frac{8\times337935-1085\times2212}{8\times167379 - 1085^2}$
$=\frac{2703480-2390020}{1339032 - 1177225}=\frac{313460}{161807}\approx1.9373$

Step3: Calculate y - intercept $b$

$b=\frac{\sum y-m\sum x}{n}=\frac{2212-1.9373\times1085}{8}=\frac{2212 - 2101.9705}{8}=\frac{110.0295}{8}=13.7537$
The line of best - fit is $y = 1.9373x+13.7537$

Step4: Estimate number of suitcases for $x = 250$

Substitute $x = 250$ into $y = 1.9373x+13.7537$
$y=1.9373\times250+13.7537=484.325+13.7537 = 498.0787\approx498$

Step1: Let $x$ be the number of years since 2012

So $x = 0$ for 2012, $x = 1$ for 2013, $\cdots$, $x = 5$ for 2017. $n = 6$
$\sum x=0 + 1+2+3+4+5 = 15$
$\sum y=340+348+356+361+375+387 = 2167$
$\sum x^2=0^2+1^2+2^2+3^2+4^2+5^2 = 55$
$\sum xy=0\times340+1\times348+2\times356+3\times361+4\times375+5\times387 = 6380$

Step2: Calculate slope $m$

$m=\frac{n\sum xy-\sum x\sum y}{n\sum x^2-(\sum x)^2}=\frac{6\times6380-15\times2167}{6\times55 - 15^2}$
$=\frac{38280-32505}{330 - 225}=\frac{5775}{105}=55$

Step3: Calculate y - intercept $b$

$b=\frac{\sum y-m\sum x}{n}=\frac{2167-55\times15}{6}=\frac{2167 - 825}{6}=\frac{1342}{6}=223.67$
The line of best - fit is $y = 55x+223.67$

Step4: For 2025, $x = 2025 - 2012=13$

Substitute $x = 13$ into $y = 55x+223.67$
$y=55\times13+223.67=715+223.67 = 938.67\approx939$

Answer:

a) $y=-2.1467x + 124.757$
b) $21.72^{\circ}F$

2.