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MAT222 Week 1 Assignment

September 22, 2014

Solving Proportions

Solving for a proportion can be used within numerous real-world problems, such as finding the population of an area. Conservationists are able to predict the population of bear’s in their area by comparing information collected from two experiments. In this problem, 50 bears in Keweenaw Peninsula were tagged and released so conservationists could estimate the bear population. One year later, the conservationist took random samples of 100 bears from the same area, proportions are able to be used in order to determine Keweenaw Peninsula’s bear 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 (Dugolpolski, 2012)?” In order to figure the estimated population, some variables need to first be defined and explain the rules for solving proportions. The ratio of originally tagged bears to the entire population is (50/x). The ratio of recaptured tagged bears to the sample size is (2/100). 50x=2100 is how the proportion is set up and is now ready to be solved. Cross multiplication is necessary for this problem. The extremes are

(100) and (50). The means are (x) and (2). 100(50)=2x New equation, and now solve for (x). 50002=2x2 Divide both sides by (2). x=2500 It is safe to assume that the bear population in Keweenaw Peninsula is estimated to be around 2500 bears.

The second problem to be completed for this assignment is problem number 10 located on page 444, in the online textbook. Problem number 10 needs to be solved for (y). A single ratio exists on both sides of the equal sign which allows the equation to be considered a proportion, and can be solved by cross multiplying the extremes and means. y-1x+3=-34

4(y-1)=-3(x+3) The result of cross multiplying the extremes and means. 4y-4= -3x-3 Distribute (4) on left side of equal sign and (-3) on the right side. 4y-4+4= -3x-3+4 Add (4) to both sides. 4y= -3x+1

4y4=-3x+14 Divide both sides by (4).

y= -3/4x+1/4 This is the linear equation, in the form of y=mx+b.

I noticed that the slope, -3/4, is the same number on the original right side of the equation, after I compared them. This tells me that there is another method that exists for solving this equation. y-1x+3=-34 y-1=-3/4(x+3) Multiply both sides of the equation by (x+3), which would cancel the denominator on the left side of the equal sign.

y-1+1= -3/4x-9/4+1 Add (1) to both sides, in order to get (y) alone, and then simplify the right side of the equal sign.

y= -3/4x+1/4 Solving the equation by the second method, would have saved me ` from completing a couple extra steps with the first method.

Neither problem solved within this assignment would be considered an extraneous solution. An extraneous solution is defined in the textbook as, “a number that appears to be a solution but causes (0) in a denominator (Dugopolski, 2012).” In order for an equation to be considered an extraneous solution, the same variable must be present in both sides of the equal sign, which neither problems offer.

Reference

Dugopolski, M. (2012). Elementary and intermediate algebra (4th ed.). New York, NY: McGraw-Hill Publishing.

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