Solving Proportions

Solving Proportions Wildlife conservations often tag members of a population in order to make estimates about things such as population size, mortality rates, mating habits, and migratory habits, and then recapture animals to see what percentage of the originally tagged groups shows up in the second sample. In this hypothetical, to estimate the size of the Keweenaw Peninsula bear population, conservationists tagged and released 50 bears. One year later, a random sample of 100 bears included only 2 tagged bears.

From these numbers the conservationists may estimate the size of the bear population using proportions based on an assumption that the tagged animals represent the average member of the targeted population. Proportions allow people to assume that the ratios remain constant across scenarios. Because proportions assume ratios are constant, they can be made equivalent to one another, allowing people to solve for a missing number if given three other numbers.

Therefore, the ratio of 2 tagged bears to 100 captured bears for the sample would be the same as the ratio of 50 total tagged bears to the unknown number of total bears for the population. The ratio of originally tagged bears to the whole population is 50/x. The ratio of recaptured tagged bears to the sample size is 2/100. Using the concept of proportions, one can make the two ratios equivalent. In other words 2/100= 50/x. The one needs to solve for x.

In this equation, 2 and x are the extremes and 50 and 100 are the means. The initial goal is to combine the extremes on one side of the equation and the means on the other side of the equation. The first step is to cross multiply, in order to get rid of the denominators. The equation that results from that cross-multiplication is 2x= 50*100 or 2x=5000. The next step is to isolate x by dividing each side by two x=5000/2. Solving for x, x=2500.

Therefore, the conservationists may estimate that the entire bear population is 2500 bears. The second math problem also uses the basic principle of proportions to equate two different numbers. The number sentence given is more complex than the ratio encountered for the bear population, but works upon the same principle. The assignment is to solve for y and identity the type of equation that results when solving for y. The number sentence given is: y-1/x+3 = -3/4.

Using cross multiplication gets rid of the denominators and yields the following equation: 4(y-1) = -3(x+3). The goal is then to isolate y on the left side of the equation, which gives one an equation solving for y. Without additional information, the equation cannot be solved for either the numerical value of x or y because doing so requires multiple equations or providing what x or y is. This information suggests immediately that the equation represents not a single set of coordinates, but a line.

As a result, one is already thinking that the end-result of the process will be an equation for a line. The slope, intercept form of an equation for a line is popularly represented as y=mx + b where m represents the slope of the line and b represents the y intercept of the line. (The y intercept is the point on the line where it intercepts the y axis; in other words, the y intercept is the point on the line where x=0).

Knowing this basic form of an equation helps guide the shaping of the equation as one solves for y. The first step.