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15 July, 14:55

Many years ago, the towns of Franklin and Chester agreed to begin posting their populations on signs just outside of their towns. They also agreed to update their signs once per year at the beginning of the year. During year 11, Franklin's sign read "/text{Franklin: Population of } 20{,}000Franklin: Population of 20,000", while Chester's sign read "/text{Chester: Population of } 25{,}000Chester: Population of 25,000". Each year, the populations grew. Specifically, Franklin's population grew by 5/%5% each year, and Chester's population grew by 500500 people each year. What is the first year in which Franklin's sign shows a larger number than Chester's sign?

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  1. 15 July, 16:29
    0
    -The first year in which Franklin's sign will show a larger number than Chester's sign will be the 9th year.

    Explanation:

    Year 1:

    -Franklin 20,000

    -Chester 25,000

    Year 2:

    -Franklin 20,000 + (20,000 x 5 / 100) : 21,000

    -Chester 25,000 + 500: 25,500

    Year 3:

    -Franklin 21,000 + (21,000 x 5 / 100) : 22,050

    -Chester 25,500 + 500: 26,000

    Year 4:

    -Franklin 22,050 + (22,050 x 5 / 100) : 23,158

    -Chester 26,000 + 500: 26,500

    Year 5:

    -Franklin 23,158 + (23,158 x 5 / 100) : 24,316

    -Chester 26,500 + 500: 27,000

    Year 6:

    -Franklin 24,318 + (24,318 x 5 / 100) : 25,534

    -Chester 27,000 + 500: 27,500

    Year 7:

    -Franklin 25,534 + (25,534 x 5 / 100) : 26,811

    -Chester 27,500 + 500: 28,000

    Year 8:

    -Franklin 26,811 + (26,811 x 5 / 100) : 28,152

    -Chester 28,000 + 500: 28,500

    Year 9:

    -Franklin 28,152 + (28,152 x 5 / 100) : 29,560

    -Chester 28,500 + 500: 29,000
  2. 15 July, 18:04
    0
    In the 9th year, the population of Franklin will exceed Chester.

    We have two equations in this problem.

    y = 20000 (1.05) ^x

    y = 25000 + 500x

    There are a few options to determine the answer.

    1) Create a chart for each city.

    2) Use a graphing calculator to graph them.

    3) Solve the equation to determine when they are equal.
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