Solving proportions Essay Example for Free

Solving proportions Essay Proportions exist in many real-world applications, and in this problem  estimating the size of the bear population on the Keweenaw Peninsula. By comparing  data from two experiments, conservationists are able to predict patterns of animal  increase or decrease. In this situation, 50 bears were captured and tagged and released to  estimate the size of the bear population. A year later, after capturing a random sample of  100 bears only 2 of the bears captured were tagged bears. These proportions will be used  to determine the bear population on the peninsula. This new bear scenario can be solved  by applying the concept of proportions which allows the assumption of the  ratio of  originally tagged bears to the whole population is equal to the ratio of recaptured tagged  bears to the size of the sample. To determine the estimated solution, the bears will be the  extraneous variables that will be defined for solving the proportions used. The ratio of originally tagged bears to the whole population X_2_The ratio of recaptured tagged bears to the sample size 10050 = _2_ This is the proportion set up and ready to solve. X 100  (50)(100), (X)(2)The next step is to cross multiply.  5000 = 2X Divide both sides by 2 2 2  2500 = XThe bear population on the Keweenaw Peninsula is estimated to be  around 2500. The extreme means for this sample were 50 and 100, X and 2.  For the second problem in this assignment, the equation must be solved for Y.  Continuing the discussion of proportions, a single fraction (ratio) exists on both sides  of the equal sign so basically it is a proportion, which can be solved by  cross  multiplying the extremes and means. Y-1 = 3 Original equation solving for Y  X+3 4  4(Y-1) = -3(X+3) Cross multiply both sides  4Y-4 = -3X-9 Add 4 to both sides  4Y = -3X-5 Divide both sides by 4  Y = -3X-5 Final answer for Y  4 4  This is a linear equation in the form of y = mx + b. After comparing the solution  to the original problem, it is noticed that the slope, -3/4 ,is the same number on the right  side of the equation. This indicates that another method exists for solving the sameequation.  Y-1 = 3 Original equation solving for Y  X+3 4  Y-1 = -3(X+3) Multiply both sides by (X+3)  4  Y-1 = -3X-9 Add 1 to both sides  4 4  Y = -3X-5 Final answer  4 4 After solving both of these problems I found it interesting how 2 totally different  equations could be solved with the same basic functions. I also found that everyday life  can incorporate these math functions to solve or estimate daily life events for a number of  different reasons.. REFERENCES References: Elementary and Intermediate Algebra, 4th Ed., Dugopolski