# Math Sl Fish Production

**Topics:**1911, 1907, 1912

**Pages:**1 (318 words)

**Published:**December 17, 2012

MODELING

The aim of this investigation is to consider commercial fishing in a particular country in two different environments, that is from the sea and a fish farm (aquaculture). The following data provided below was taken form the UN Statistics Division Common Database. The tables gives the total mass of fish caught in the sea, in thousands of tones (1 tone = 1000 kilograms). Year| 1980| 1981| 1982| 1983| 1984| 1985| 1986| 1987| 1988| Total Mass| 426.8| 470.2| 503.4| 557.3| 564.7| 575.4| 579.8| 624.7 | 669.9|

Year| 1989| 1990| 1991| 1992| 1993| 1994| 1995| 1996| 1997| Total Mass| 450.5| 379.0| 356.9| 447.5| 548.8| 589.8| 634.0| 527.8 | 459.1|

Year| 1998| 1999| 2000| 2001| 2002| 2003| 2004| 2005| 2006| Total Mass| 487.2| 573.8| 503.3| 527.7| 566.7| 507.8| 550.5| 426.5 | 533.0| Table1. – The total mass of fish caught from 1980-2006

From the presented data, it is shown that in some years the amount of fish caught rises and on others the amount of fish caught in the sea decreases. The reason of the amount of fish to decrease can be influenced by ecological factors and weather phenomena’s or even the amount of fishermen on the particular area in the sea. Identifying variables: let y stand for the amount of fish caught from the sea and x stands for the years passed. It shows yx=amount of fish caught, so e.g. y1980=426.8. The set of possible values of the variable on the horizontal axis is called the domain. Therefore {x | 1980 < x ≤ 2006} is the domain of the amount of fish caught. The set which describes the possible y-values is called the range. Therefore the range of the fish caught is the total mass of fish from the sea.

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