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# Mat222 Week 1 Assignment

Submitted By caboun6828
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Solving Proportions To estimate the size of the bear population on Keweenaw Peninsula, conservationists Captured, tagged, and released 50 bears. One year later, a random sample of 100 bears included only 2 tagged bears. The question I am asked is what do conservationist’s estimate of the size of the bear population? You will notice while reading question #56 on page 437, we are to assume that the ratio of originally tagged bears to the whole population is equal to the ratio of recaptured tagged bears to the size of the sample. The ratio of originally tagged bears to the whole population is 50X The ration of recaptured tagged bears to the sample size is 2100
50X=2100 This is the proportion set up and ready to solve. I will cross multiply setting the extremes equal to the means.
10050=2x 100and 50 are the extremes, while X and 2 are the means.
50002=2x2 Divide both sides by 2
X = 2500 The bear population on the Keweenaw Peninsula is around 2500 bears. The second problem for assignment one week one I am asked to solve the below equation for y. The first thing I notice is that 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 the means. y-1x+3=-34 Is the equation I am asked to solve.
3y-1=-3x-4 The result of the cross multiplying.
3y-3=-3x+12 Distribute 3 on the left side and -3 on the right.
3y-3+3=-3x+12+3 Subtract 3 from both sides.
3y=-3x+15
3y3=-3x3+153 Divide both sides by 3 y=-x+5 This is a linear equation in the form of y=mx+b. This equation is in its simplest form. I like how we can take just a couple of numbers from a word equation and put it in an order that will help us solve many estimates we may come across.

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