Question

unit 1 do nows
do now 1 - 1.1 function basics

1. in which quadrant do lines l and m intersect?

name brianna moore

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	422
1984	432
1988	404
1992	420
1994	395
1998	382

a) find the line of best fit:
b) estimate the 500 - meter time for the 2020 olympics.
c) write a description of data:
c) write a description of data:
c) write a description of data:
c) write a description of data:

Explanation:

Step1: Recall the formula for the line of best - fit (least - squares regression line)

The line of best - fit for a set of data points \((x_i,y_i)\) is of the form \(y = mx + b\), where \(m=\frac{n\sum_{i = 1}^{n}x_iy_i-\sum_{i = 1}^{n}x_i\sum_{i = 1}^{n}y_i}{n\sum_{i = 1}^{n}x_i^{2}-(\sum_{i = 1}^{n}x_i)^{2}}\) and \(b=\overline{y}-m\overline{x}\), \(\overline{x}=\frac{1}{n}\sum_{i = 1}^{n}x_i\), \(\overline{y}=\frac{1}{n}\sum_{i = 1}^{n}y_i\), and \(n\) is the number of data points.

Step2: For the first data set (latitude - temperature)

Let \(x\) be the latitude and \(y\) be the temperature.
1. Calculate \(n = 9\), \(\sum_{i=1}^{9}x_i=0 + 10+20+\cdots+80 = 360\), \(\sum_{i = 1}^{9}y_i=79 + 81+\cdots+1=392\), \(\sum_{i=1}^{9}x_i^{2}=0^{2}+10^{2}+\cdots+80^{2}=20400\), \(\sum_{i = 1}^{9}x_iy_i=0\times79+10\times81+\cdots+80\times1 = 1470\).
2. Calculate \(m=\frac{9\times1470 - 360\times392}{9\times20400-(360)^{2}}\)
\[ \begin{align*} m&=\frac{13230-141120}{183600 - 129600}\\ &=\frac{- 127890}{54000}\\ &=- 2.3683 \end{align*} \]
3. Calculate \(\overline{x}=\frac{360}{9}=40\), \(\overline{y}=\frac{392}{9}\approx43.56\)
4. Calculate \(b = 43.56-(-2.3683)\times40=43.56 + 94.732=138.292\)
The line of best - fit is \(y=-2.3683x + 138.292\)
5. For \(x = 48\), \(y=-2.3683\times48+138.292=-113.6784 + 138.292 = 24.6136\approx24.61\)

Step3: For the second data set (passengers - suitcases)

Let \(x\) be the number of passengers and \(y\) be the number of suitcases.
1. Calculate \(n = 8\), \(\sum_{i = 1}^{8}x_i=75 + 92+\cdots+200 = 1085\), \(\sum_{i = 1}^{8}y_i=159+180+\cdots+405 = 1922\), \(\sum_{i = 1}^{8}x_i^{2}=75^{2}+92^{2}+\cdots+200^{2}=177933\), \(\sum_{i = 1}^{8}x_iy_i=75\times159+92\times180+\cdots+200\times405 = 303995\)
2. Calculate \(m=\frac{8\times303995-1085\times1922}{8\times177933-(1085)^{2}}\)
\[ \begin{align*} m&=\frac{2431960 - 2085370}{1423464-1177225}\\ &=\frac{346590}{246239}\\ &\approx1.4075 \end{align*} \]
3. Calculate \(\overline{x}=\frac{1085}{8}=135.625\), \(\overline{y}=\frac{1922}{8}=240.25\)
4. Calculate \(b = 240.25-1.4075\times135.625=240.25 - 190.877=49.373\)
The line of best - fit is \(y = 1.4075x+49.373\)
5. For \(x = 250\), \(y=1.4075\times250+49.373=351.875+49.373 = 401.248\approx401.25\)

Step4: For the third data set (year - graduates)

Let \(x\) be the number of years since 2012 (\(x = 0\) for 2012, \(x = 1\) for 2013, etc.).
1. Calculate \(n = 6\), \(\sum_{i = 1}^{6}x_i=0 + 1+\cdots+5 = 15\), \(\sum_{i = 1}^{6}y_i=340+348+\cdots+387 = 2167\), \(\sum_{i = 1}^{6}x_i^{2}=0^{2}+1^{2}+\cdots+5^{2}=55\), \(\sum_{i = 1}^{6}x_iy_i=0\times340+1\times348+\cdots+5\times387 = 5915\)
2. Calculate \(m=\frac{6\times5915-15\times2167}{6\times55-(15)^{2}}\)
\[ \begin{align*} m&=\frac{35490-32505}{330 - 225}\\ &=\frac{2985}{105}\\ &=28.4286 \end{align*} \]
3. Calculate \(\overline{x}=\frac{15}{6}=2.5\), \(\overline{y}=\frac{2167}{6}\approx361.17\)
4. Calculate \(b = 361.17-28.4286\times2.5=361.17 - 71.0715=290.0985\)
The line of best - fit is \(y = 28.4286x+290.0985\)
5. For \(x = 13\) (2025 - 2012), \(y=28.4286\times13+290.0985=369.5718+290.0985 = 659.6703\approx659.67\)

Step5: For the fourth data set (year - time)

Let \(x\) be the number of years since 1980 (\(x = 0\) for 1980, \(x = 4\) for 1984, etc.).
1. Calculate \(n = 6\), \(\sum_{i = 1}^{6}x_i=0 + 4+\cdots+18 = 44\), \(\sum_{i = 1}^{6}y_i=422+432+\cdots+382 = 2455\), \(\sum_{i = 1}^{6}x_i^{2}=0^{2}+4^{2}+\cdots+18^{2}=580\), \(\sum_{i = 1}^{6}x_iy_i=0\times422+4\times432+\cdots+18\times382 = 16778\)
2. Calculate \(m=\frac{6\times16778-44\times2455}{6\times580-(44)^{2}}\)
\[ \begin{align*} m&=\frac{100668-107020}{3480 - 1936}\\ &=\frac{- 6352}{1544}\\ &=- 4.114 \end{align*} \]
3. Calculate \(\overline{x}=\frac{44}{6}\approx7.33\), \(\overline{y}=\frac{2455}{6}\approx409.17\)
4. Calculate \(b = 409.17-(-4.114)\times7.33=409.17+30.1556 = 439.3256\)
The line of best - fit is \(y=-4.114x + 439.3256\)
5. For \(x = 40\) (2020 - 1980), \(y=-4.114\times40+439.3256=-164.56+439.3256 = 274.7656\approx274.77\)

Answer:

1.
a) \(y=-2.3683x + 138.292\)
b) \(24.61\)
c) As the latitude increases, the average temperature decreases approximately linearly.
2.
a) \(y = 1.4075x+49.373\)
b) \(401.25\)
c) The number of suitcases increases approximately linearly with the number of passengers.
3.
a) \(y = 28.4286x+290.0985\)
b) \(659.67\)
c) The number of graduating seniors has been increasing approximately linearly since 2012.
4.
a) \(y=-4.114x + 439.3256\)
b) \(274.77\)
c) The Olympic 500 - meter Gold Medal Speed Skating times have been decreasing approximately linearly over the years.

Answer:

1.
a) \(y=-2.3683x + 138.292\)
b) \(24.61\)
c) As the latitude increases, the average temperature decreases approximately linearly.
2.
a) \(y = 1.4075x+49.373\)
b) \(401.25\)
c) The number of suitcases increases approximately linearly with the number of passengers.
3.
a) \(y = 28.4286x+290.0985\)
b) \(659.67\)
c) The number of graduating seniors has been increasing approximately linearly since 2012.
4.
a) \(y=-4.114x + 439.3256\)
b) \(274.77\)
c) The Olympic 500 - meter Gold Medal Speed Skating times have been decreasing approximately linearly over the years.