Solving Proportions Paper

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Solving Proportions MAT 222 October 06, 2013 Solving Proportions Proportions are used in everyday life. In the first problem that I am going to do we will be finding out the estimated bear population with the information given to us from Elementary and Intermediate Algebra page 437 problem 56. According to the textbook the idea of proportions is to show the equality of two ratios (Dugolpolski, 2012, pg. 431, sec. 6.7). The instructions are to set up the two ratios and write the equation, using an appropriate variable for the population. In order to determine the estimated population I will be defining variables and rules for solving this problem. Page 437 problem 56 is as follows; Bear population. To estimate the size of…show more content…
The ratio of the tagged bears to the random sample size is 48/150. 300 = 48 This is the proportion set up which makes it ready to solve. Cross multiplication X 150 is necessary at this point. The extremes are 300 and 150 and the means are x and 48. 150(300) = 48x 45000 = 48x Now we need to divide both sides 48. 48 48 X = 1000 The estimated size of the bear population on the Keweenaw Peninsula is estimated to be around 1000 bears. Now I am to solve the second problem out of Elementary and Intermediate Algebra problem 10 on page 444 for this assignment. The problem is as follows; y – 1/x +3 = -3/4 It appears that there is a single fraction ratio on both sides of the equal sign which means that this problem is a proportion. This means that this problem can be solved cross multiplying the extremes and means of the problem. y – 1/x +3 = -3/4 This is the problem that we were given to solve for y. First we multiply both sides by x+3 so we will have y-1(x+3) =-3/4(x+3) Then we simplify so we are left with y-1= -3/4x+3 Then we add 1 to both sides which gives us y-1(+1) = -3/4x+ 3(+1) Now we simplify again which us leaves
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