# Slove for X

Solving a proportion as we learned in this week reading, means that we are missing an import number in our equation or fraction, and we need to solve for that missing value. For the first proportion, number 56 on page 437 of Elementary and Intermediate Algebra by Dugopolski: 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? Beginning to solve this equation first, we needed to make a proportion with the number of tagged bears in the sample and in the population
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? Let "p" be the number in the population. Real population / sample caught = released population / sample caught.
To estimate the bear population, set up a proportion, which is just two ratios that are equal.

2         50

----   =   ----

100         P

Here P is the value of the bear population we are trying to estimate.   To find P, the estimated bear population, cross multiply the extremes (2 and P) and the means (100 and 50) of the proportion:

2P = 50*100

P = 2500 bears
50002=2x2 Divide both sides by 2
P=2500The bear population of Keweenaw Peninsula is estimated to be around 2500.   I don’t see an   extraneous solutions.
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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 means.

This is the proportion set up and ready to solve. I will cross multiply the x...