Mat 222 Week 1 Assignment
Solving Proportions
Tara Lint
MAT 222 Week 1 Assignment
Instructor: James Segala
August 18, 2013
Solving Proportions
Proportions exist in the real world. For example, in finding out the price of a unit, or the population of a specific species. The first problem that we are working with states that “. Bear population. 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?(Dugolpolski, 2012)
In reading over the “Bear Population” #56 on page 437 (Dugolpolski, 2012), the concept of proportions allow the assumption 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 estimated solution, variables will be defined and rules for solving proportions are used.
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.
50=2 This is the proportion set up and ready to solve. This is the step where we will cross multiply. x 100 at this point. The extremes are 100 and 50. The means are x and 2
100(50)=2x
50002=2x2 Divide both sides by 2
X=2500 The bear population of Keweenaw Peninsula is estimated to be around 2500.
For the second problem of this assignment the equation must be solved for y. Therefore, by continuing proportions, a single ratio (fraction) exist on both sides of the equal sign. Therefore, it is a proportion, which is solved by cross multiplying the extreme means. So in this problem we are working problem #10 on page 444 (Dugolpolski, 2012). y-1x+3=-34 The original equation y-1=-3 x+3 4
4(y-1)=-3(x+3) The result of cross multiplying
4y-4=-3x-9 Distribute 4 on the left side and -3 on the right side
4y-4+4=-x+3+4 Add 4 to both sides and
Tara Lint
MAT 222 Week 1 Assignment
Instructor: James Segala
August 18, 2013
Solving Proportions
Proportions exist in the real world. For example, in finding out the price of a unit, or the population of a specific species. The first problem that we are working with states that “. Bear population. 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?(Dugolpolski, 2012)
In reading over the “Bear Population” #56 on page 437 (Dugolpolski, 2012), the concept of proportions allow the assumption 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 estimated solution, variables will be defined and rules for solving proportions are used.
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.
50=2 This is the proportion set up and ready to solve. This is the step where we will cross multiply. x 100 at this point. The extremes are 100 and 50. The means are x and 2
100(50)=2x
50002=2x2 Divide both sides by 2
X=2500 The bear population of Keweenaw Peninsula is estimated to be around 2500.
For the second problem of this assignment the equation must be solved for y. Therefore, by continuing proportions, a single ratio (fraction) exist on both sides of the equal sign. Therefore, it is a proportion, which is solved by cross multiplying the extreme means. So in this problem we are working problem #10 on page 444 (Dugolpolski, 2012). y-1x+3=-34 The original equation y-1=-3 x+3 4
4(y-1)=-3(x+3) The result of cross multiplying
4y-4=-3x-9 Distribute 4 on the left side and -3 on the right side
4y-4+4=-x+3+4 Add 4 to both sides and