# Math Equations

My Mathematical Equations

By Kathleen Rossi

MAT 222 Intermediate Algebra

Instructor: Mohamed Elseifieen

May 12, 2013

My Mathematical Equations

This paper will show two mathematical problems, the first is “To estimate the size of the bear population on the Keweenaw Peninsula, conservationists captured, tagged, and released 50 bears. One year later, a random sample of 100 bears included only 2 tagged bears. What is the conservationist's estimate of the size of the bear population?” (Dugopolski, 2013, pp. 437, probem 56). The second will be to complete problem 10 on page 444 of Elementary and Intermediate Algebra. Here all steps in solving the problem will be explained step by step. The first problem is to estimate the size of the bear population located on the Keweenaw Peninsula conservation. In reading over the “Bear Population” method #56 on page 437you will notice we are to assume that the ratio of originally tagged bears to the whole population is equal to the ratio of recaptured bears to the size of the sample. The ratio of the originally tagged bears to the whole population is 2100 The ration of the recaptured tagged bears to the sample size is 50x 2100=50x Since x is on the right-hand side of the equation, we need to switch the sides so it is on the left-hand side. 50x=2100 This is the proportion set up and ready to solve. I will cross multiply setting the extremes equal to the means. 100(50) = 2x Here 100 and 50 are the extreme, while x and 2 are the means. 50002=2x2 Next we must divide each term in the equation by 2. X=50002 Cancel out the common factor

X=2500 The bear population in Keweenaw Peninsula is estimated to be around 2500. For the second problem in this assignment I am asked to solve this equation for y. The first thing I notice is that it is a single fraction (ratio) on both sides of the equal sign so basically it is a proportion which can be solved by cross multiplying the extremes and...

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