Solving proportions
Solving proportions
Proportions exist in many realworld 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.
Y1 = – 3 Original equation solving for Y X+3 4 4(Y1) = 3(X+3) Cross multiply both sides 4Y4 = 3X9 Add 4 to both sides 4Y = 3X5 Divide both sides by 4 Y = 3X5 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. Y1 = – 3 Original equation solving for Y X+3 4 Y1 = 3(X+3) Multiply both sides by (X+3) 4 Y1 = 3X9 Add 1 to both sides 4 4 Y = 3X5 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
A

Subject: Assignment, Education, Science,

University/College: University of California

Type of paper: Thesis/Dissertation Chapter

Date: 6 March 2016

Words:

Pages:
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