Solving Proportions
Solving Proportions
MAT 222
Instructor: Dr. Stacie Williams
Solving Proportions
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? Let’s make "x" the number of the total bear population.
The ratio of originally tagged bears to the whole population is 50x
The ratio of the random tagged bears of bears to the random sample of bears is 2100
50x = 2100 This is the proportion that will be set up and ready to solve. I will then cross multiply setting the extremes equal to the means.
50(100)=2(x) 50 and 100 are the extremes while x and 2 are the means.
50002= 2x2 Divide both sides by 2 x =2500 bears The bear population of the Keweenaw Peninsula is estimated to be around 2500 bears
Tagged by conservationists = 50 bears which represented an estimated 2% of the population 50/.02=2500 total bear population.
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-1x+3= - 34
4y-1= -3(x+3) The result of the cross multiplying.
4y-4= -3x-9 Distribute 4 on the left and -3 on the right.
4y-4+4= -3x-9+4 Add 4 to both sides.
4y= -3x-5
4y4= -3x4- 54 Divide both sides by 4
y= -34x- 54 This linear equation is in the form of y = mx + b.
There is also another way to solve this proportion. The equation could also be solved if (x+3) was multiplied on both sides which cancels out the denominator on the left.
(x+3)y-1x+3= - 3(x+3)4
Then you would add 1 to both sides to isolate the y and simplify the right side. This process would eliminate a few steps from the traditional
MAT 222
Instructor: Dr. Stacie Williams
Solving Proportions
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? Let’s make "x" the number of the total bear population.
The ratio of originally tagged bears to the whole population is 50x
The ratio of the random tagged bears of bears to the random sample of bears is 2100
50x = 2100 This is the proportion that will be set up and ready to solve. I will then cross multiply setting the extremes equal to the means.
50(100)=2(x) 50 and 100 are the extremes while x and 2 are the means.
50002= 2x2 Divide both sides by 2 x =2500 bears The bear population of the Keweenaw Peninsula is estimated to be around 2500 bears
Tagged by conservationists = 50 bears which represented an estimated 2% of the population 50/.02=2500 total bear population.
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-1x+3= - 34
4y-1= -3(x+3) The result of the cross multiplying.
4y-4= -3x-9 Distribute 4 on the left and -3 on the right.
4y-4+4= -3x-9+4 Add 4 to both sides.
4y= -3x-5
4y4= -3x4- 54 Divide both sides by 4
y= -34x- 54 This linear equation is in the form of y = mx + b.
There is also another way to solve this proportion. The equation could also be solved if (x+3) was multiplied on both sides which cancels out the denominator on the left.
(x+3)y-1x+3= - 3(x+3)4
Then you would add 1 to both sides to isolate the y and simplify the right side. This process would eliminate a few steps from the traditional