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3 years ago

Options- -4,2 1/2

Mathematics

1 answer:

nirvana33 [79]3 years ago

5 0

Answer:

Q=(4,2 1/2)

Step-by-step explanation:

In graph, upper right hand side cointain positive sign i.e + + and Q also located there.so...

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liraira [26]

If we see the data closely, a pattern emerges. The pattern is that the ratio of the population of every consecutive year to the present year is 1.6

Let us check it using a couple of examples.

The rabbit population in the year 2010 is 50. The population increases to 80 the next year (2011). Now, $\frac{80}{50}=1.6$

Likewise, the rabbit population in the year 2011 is 80. The population increases to 128 the next year (2012). Again, $\frac{128}{80}$ $=1.6$

We can verify the same ratio with all the data provided.

Thus, we know that the population in any given year is 1.6 times the population of the previous year. This is a classic case of a compounding problem. We know that the formula for compounding is as:

$F=P\times r^n$

Where $F$ is the future value of the rabbit population in any given year

$P$ is the rabbit population in the year "0" (that is the starting year 2010) and that is 50 in this question. (please note that there is just one starting year).

$r$ is the ratio multiple with which the rabbit population increases each consecutive year.

$n$ is the nth year from the start.

Let us take an example for the better understanding of the working of this formula.

Let us take the year 2014. This is the 4th year

So, the rabbit population in 2014 should be:

$F_{2014} =50\times(1.6)^4\approx328$

This is exactly what we get from the table too.

Thus, $F=P\times r^n$ aptly represents the formula that dictates the rabbit population in the present question.

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