the table shows the population of center city in various years. use the data from 1990 and 2005 to create a…
the table shows the population of center city in various years. use the data from 1990 and 2005 to create a linear model that predicts the population of the city (y) in a given year (x). in which year was the actual population of center city most different from the value predicted by this model?\n\n| year | city population |\n| ---- | ---- |\n| 1985 | 194,957 |\n| 1990 | 197,800 |\n| 1992 | 199,532 |\n| 2000 | 203,750 |\n| 2005 | 206,561 |\n| 2012 | 210,600 |\n\no 1985\no 1992\no 2000\no 2012
Answer
Explanation:
Step1: Define variables
Let $x$ be the number of years since 1990. So for 1990, $x = 0$; for 2005, $x=2005 - 1990=15$. Let $y$ be the population. The two - point form of a line is $y - y_1=\frac{y_2 - y_1}{x_2 - x_1}(x - x_1)$. We have $(x_1,y_1)=(0,197800)$ and $(x_2,y_2)=(15,206561)$.
Step2: Calculate the slope
The slope $m=\frac{y_2 - y_1}{x_2 - x_1}=\frac{206561 - 197800}{15}=\frac{8761}{15}\approx584.07$. The equation of the line in slope - intercept form $y=mx + b$. Since $b = y_1=197800$ when $x_1 = 0$, the linear model is $y = 584.07x+197800$.
Step3: Calculate predicted values for each year
For 1985 ($x=- 5$): $y=584.07\times(-5)+197800=-2920.35 + 197800=194879.65$. The difference from the actual value $|194957 - 194879.65| = 77.35$. For 1992 ($x = 2$): $y=584.07\times2+197800=1168.14+197800=198968.14$. The difference from the actual value $|199532 - 198968.14|=563.86$. For 2000 ($x = 10$): $y=584.07\times10+197800=5840.7+197800=203640.7$. The difference from the actual value $|203750 - 203640.7| = 109.3$. For 2012 ($x = 22$): $y=584.07\times22+197800=12849.54+197800=210649.54$. The difference from the actual value $|210600 - 210649.54| = 49.54$.
Answer:
1992