1. the data in the table below shows the average temperature in northern latitudes:\nlatitude (°n) 0 10 20…

1.

Explanation:

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

Answer:

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

2.

Explanation:

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

Answer:

a) $y = 1.9373x+13.7537$ b) $498$

3.

Explanation:

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) $y = 55x+223.67$ b) $939$

4.

Explanation:

Step1: Let $x$ be the number of years since 1980

So $x = 0$ for 1980, $x = 4$ for 1984, $\cdots$, $x = 18$ for 1998. $n = 6$ $\sum x=0 + 4+8+12+14+18 = 56$ $\sum y=422+432+404+420+395+382 = 2455$ $\sum x^2=0^2+4^2+8^2+12^2+14^2+18^2 = 888$ $\sum xy=0\times422+4\times432+8\times404+12\times420+14\times395+18\times382 = 21478$

Step2: Calculate slope $m$

$m=\frac{n\sum xy-\sum x\sum y}{n\sum x^2-(\sum x)^2}=\frac{6\times21478-56\times2455}{6\times888 - 56^2}$ $=\frac{128868-137480}{5328 - 3136}=\frac{-8612}{2192}\approx - 3.93$

Step3: Calculate y - intercept $b$

$b=\frac{\sum y-m\sum x}{n}=\frac{2455-(-3.93)\times56}{6}=\frac{2455 + 220.08}{6}=\frac{2675.08}{6}=445.85$ The line of best - fit is $y=-3.93x + 445.85$

Step4: For 2020, $x = 2020 - 1980 = 40$

Substitute $x = 40$ into $y=-3.93x + 445.85$ $y=-3.93\times40+445.85=-157.2+445.85 = 288.65$

Answer:

a) $y=-3.93x + 445.85$ b) $288.65$ s