A researcher determined the amount of ducks that were in a pond over a four-year time span. The results are shown in the table below.

Question

A researcher determined the amount of ducks that were in a pond over a four-year time span. The results are shown in the table below.
Year 1, 2, 3, 4
Ducks 64, 78, 92. 106

Year Ducks
1 64
2 78
3 92
4 106
Select an equation that best models the amount of ducks (d) that will be in the pond during the nth year.

d= 64(1.22)^n-1

d=14n +64

d= 14n + 50

d + 50(1.14)^n-1

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Julia 3 days 2021-09-15T03:47:14+00:00 2 Answers 0

Answers ( )

    0
    2021-09-15T03:48:16+00:00

    Answer:

    Option C (d = 14n + 50).

    Step-by-step explanation:

    The observations for Year 1, 2, 3, and 4 are 64, 78, 92, and 106 respectively. It can be observed that the difference between second term and the first term is 78 – 64 = 14. This is also true for the third term and the second term. In fact, the difference between the terms is fixed. This means that the sequence is an arithmetic sequence, in which the difference between the terms is fixes for all the terms. The nth term in this case is defined as:

    S = a + (n-1)*d; where S is the output at a given n, a is the first term, n is the position, and d is the common difference. In this question, S = ducks (d), a = 64, and d = 14. Therefore, plugging in the values in the above formula gives:

    d = 64 + (n-1)*14.

    d = 64 + 14n – 14.

    d = 14n + 50.

    Therefore, the model which best describes the amount of ducks that will be in the pond during the nth year is d = 14n + 50, i.e. Option C!!!

    0
    2021-09-15T03:48:20+00:00

    Answer: Third Option

    a_n =14n+50

    Step-by-step explanation:

    By subtracting consecutive terms from the number of ducks each year you will get a common difference of 144.

    78-64 = 14\\92-78 = 14\\106-92 = 14

    This means that the data can be modeled by the equation of an arithmetic sequence

    The arithmetic sequences have the following form:

    a_n = a_1 + d(n-1)

    Where a_1 is the first term and d is the common difference between the consecutive terms.

    So :

    a_1 = 64

    d = 14

    Then the equation is:

    a_n = 64 + 14(n-1)

    a_n = 64 + 14n-14

    a_n =14n+50

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