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23 March, 18:47

There were 200,000 animals of a certain species in 1980. Since then, this number has decreased by 4.5% each year. Approximately how many animals of this species will be left in 2025?

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  1. 23 March, 19:05
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    There'll be approximately 25187.3059 animals of this species in 2025.

    Step-by-step explanation:

    In this case we have a compounded interest problem, but the interest rate is negative, since the number will be decreasing. To solve it we can use the compound interest formula shown bellow:

    M = C * (1 + r) ^ (t)

    Where M is the final amount, C is the initial amount, r is the interest rate and t is the time elapsed.

    In this case the time elapsed was from 1980 to 2025, so 45 years. Applying the data to the formula gives us:

    M = 200000 * (1 + (-0.045)) ^ (45)

    M = 200000 * (1 - 0.045) ^ (45)

    M = 200000 * (0.955) ^ (45)

    M = 25187.3059

    There'll be approximately 25187.3059 animals of this species in 2025.
  2. 23 March, 19:13
    0
    Answer: Approximately 25187 animals of this species will be left in 2025

    Step-by-step explanation:

    We would apply the formula for exponential decay which is expressed as

    y = b (1 - r) ^x

    Where

    y represents the population of animals after x years.

    x represents the number of years.

    b represents the initial population of animals.

    r represents rate of decay.

    From the information given,

    b = 200000

    r = 4.5% = 4.5/100 = 0.045

    x = 2025 - 1980 = 45 years

    Therefore,

    y = 200000 (1 - 0.045) ^45

    y = 200000 (0.955) ^45

    y = 25187
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