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Solving Proportions

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SOLVING PROPORTIONS
Melissa Helm
MAT 222
Instructor Jennie Strong
October 27, 2014

For this week’s assignment, we were given two problems to solve.
The first one is problem #56 from page 437.
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 “x” be the number we will find for the bear population. x = 100 I will cross multiply this equation.
50 2
2x = 50 * 100 After cross multiplying.
2x = 5,000 Multiplied the number of released bears to the Sample bears
X = 2,500 Divided 2 on each side and this is the final answer Of how many bears is the estimated population.
The second problem is #10 from page 444.
For the second equation, I need to solve for y. This is a single fraction ratio on both sides of the equal sign. This would be considered a proportion which I can solve by cross multiplying the extreme means of the equation. y-1 = -3 Equation x+3 4 y-1 * (x+3) = -3 * (x+3) Multiply x+3 to both sides. x+3 4 y-1 = -3x+3 Add 1 to 3. This appears to be a solution but causes 4. 0 in a denominator is called an extraneous Solution.
Y = -3x +4 Answer. 4
The type of equation that was the answer in problem 10 is called a linear equation. The coefficient of x is different than the original problem. It was x+3 and in this case, it was -3x/4.

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