Question

1. in 1960, a certain city has population 750,000. in 2005, that same city now has a population of 1,200,000. find a linear equation model (y = m(x - h)+b) that fits this data. then predict the population for the years 1984 and 2025. predict the years that the population will hit 1,500,000 and 2,000,000.

Explanation:

Step1: Calculate the slope \(m\)

Let \(x = 0\) represent the year 1960 (\(h = 0\)), so \(b=750000\). For the year 2005 (\(x = 2005 - 1960=45\)), \(y = 1200000\).
The slope formula is \(m=\frac{y - b}{x - h}\). Substituting the values: \(m=\frac{1200000 - 750000}{45-0}=\frac{450000}{45}=10000\)

Step2: Write the linear equation

Using the point - slope form \(y=m(x - h)+b\), with \(h = 0\), \(m = 10000\) and \(b = 750000\), the equation is \(y = 10000x+750000\)

Step3: Predict the population for 1984

For 1984, \(x=1984 - 1960 = 24\). Substitute \(x = 24\) into \(y = 10000x+750000\): \(y=10000\times24 + 750000=240000+750000 = 990000\)

Step4: Predict the population for 2025

For 2025, \(x=2025 - 1960=65\). Substitute \(x = 65\) into \(y = 10000x+750000\): \(y=10000\times65+750000=650000 + 750000=1400000\)

Step5: Find the year when \(y = 1500000\)

Set \(y = 1500000\) in \(y = 10000x+750000\). Then \(1500000=10000x+750000\).
Subtract 750000 from both sides: \(10000x=1500000 - 750000=750000\).
Divide by 10000: \(x=\frac{750000}{10000}=75\). The year is \(1960 + 75=2035\)

Step6: Find the year when \(y = 2000000\)

Set \(y = 2000000\) in \(y = 10000x+750000\). Then \(2000000=10000x+750000\).
Subtract 750000 from both sides: \(10000x=2000000 - 750000 = 1250000\).
Divide by 10000: \(x=\frac{1250000}{10000}=125\). The year is \(1960+125 = 2085\)

Answer:

• Linear equation: \(y = 10000x+750000\)
• Population in 1984: \(990000\)
• Population in 2025: \(1400000\)
• Year when population is \(1500000\): \(2035\)
• Year when population is \(2000000\): \(2085\)