Question

In 1960, the population of a town was 14 thousand people. Over the course of the next 50 years, the town grew at a rate of 30 people per year. Hint: let t=0 be 1960, and t-1 be 1961, etc. A) Assuming this continues, what is the population predicted to be in 2030? B) Set up and solve the equation to find in which year the population will reach 15 thousand. Give your answer in a form like 1980 or 1994. A) B) people​

Answer 1

`\textbf {Allow me to assist you.}`

`\textrm {Let's understand the values we are given :-}`

• 
  `\textrm {Starting year = 1960}`
• 
  `\textrm {Starting population = 14,000}`
• 
  `\textrm {Rate of Growth = 30 people/year}`

`\textrm {Since a fixed number of people increase each year,}\\\textrm {this can be considered as a linear equation :}\\\fbox {y = mx + c}`

`\textrm {Then, our equation is :}`

• 
  `\textrm {y = mx + c}`
• 
  `\textrm {y = 30x + 14,000}`

`\textrm {Using this equation, let us solve for parts (A) and (B).}`

`\underline {Part(A)}`

`\textrm {Find the number of years after 1960.}`

• 
  `\textrm {x = 2030 - 1960}`
• 
  `\textrm {x = 70}`

`\textrm {Now, substitute in the equation.}`

• 
  `\textrm {y = 30(70) + 14,000}`
• 
  `\textrm {y = 2,100 + 14,000}`
• 
  `\fbox {y = 16,100}`

`\underline {Part(B)}`

`\textrm {Let y = 15,000.}`

• 
  `\mathrm {15,000 = 30x + 14,000}`
• 
  `\textrm {30x = 1,000}`
• 
  `\mathrm {x = (100)/(3) = 33.3}`

`\textrm {Round to the lower nearest integer, which is 33.}`

• 
  `\textrm {Year in which it will reach : 1960 + 33}`
• 
  `\fbox {1993}`