# the size of the sample.

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. 37, 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. 0002=2×2 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 means. y-1)=-34*((x+3))Multiply both sides of the equation by (x+3) (y-1)=-34*(x+3)Remove the extra parentheses (y-1) = -3(x+3)4Multiply the rational expressions to get – 3(x+3)4 y-1= -3(x+3)4Remove the parentheses around the expression y-1 Since -1 does not contain the variable solve for, move it to the right-hand side of the equation by adding 1 to both sides. y=1 *44 – 3(x+3)4Multiply each term by a factor of 1 that will equate all the denominators. In this case, all terms need a denominator of 4. = 14 (-1*4) 3(x+3)4Multiply the expression by a factor of 1 to create the least common denominator (LCD) of 4. y= 44 – 3(x+3)4 Multiply 1 by 4 to get 4. y = 4-3(x+3)4

The numerators of expressions that have equal denominators can be combined. In this case, (4)4 and – 3(x+3)4 have the same denominator of 4, so the numerators can be combined. y = 4-3 (x+3)4 Remove the parentheses around the expression 4. y = 4-3 x-3 (3)4 Multiply -3 by each term inside the parentheses (x-3). = 4-3 *x-3 (3)4 Multiply -3 by the x inside the parentheses. y = 4-3 x-3 (3)4 Multiply -3 by x to get – 3x. y = 4-3 x-3* 34 Multiply -3 by the 3 inside the parentheses. y = 14 (4-3x-9) Multiply -3 by 3 to get -9. y = 14 (-5 -3x) Subtract 9 from 4 to get -5 y = 14 (-3x -5) Reorder the polynomial -5 -3x alphabetically from left to right, starting with the highest order term. y = 14 (-3x-5)Simplify the right-hand side of the equation.

Here I have notice that the solution for y = 14 (-3x-5) but the equation for x is different and x as a rational expression is (y-1) · 4=-3 · (x+3) which is solved by which is solved by a ration. Simplify the equation for x = – 4y3 – 53 . Another thing I have learned is if you go from x = y to x^2 = y^2 you have introduced the “extraneous” root. References: Dugopolski, M. (2013). Elementary and Intermediate Algebra, 4th Ed. New York, NY.