Body Mass Index Jacob Shearer MAT221 Introduction to Algebra Instructor Pamela Clarke November 17, 2013 I am going to solve the following 4 formulas involving Body Mass Index starting with the first one. 17<703WH^2<22 This is the equivalent inequality where I substituted BMI with the formula. Then I replace the H^2 with my height in inches. 17<703W73^2<22 After factoring in my height to the second power I get 17<703W5329<22 Now I multiply all 3 terms by the denominator 17(5329) < 703W(5329)5329 <22(5329) and then get 90593 < W < 117238. Then finally I divide all 3 by 703 in order to isolate W. 128.8 < W < 166.7 So people with the height of 73 inches could have a longer than average life span if they weigh between 129 and 167 pounds.
500.1 grams o B. 501.0 grams o C. 510 grams o D. 600 grams 3. 3. You are determining the mass of an object using a triple beam balance. When the pointer is lined up with the zero mark, the riders show values of 300 grams, 20 grams, and 8.0 grams.
INEQUALITIES 1 BMI Brandon Picone MAT 221: Introduction to Algebra Instructor: Gregory Dlabach Dec 2, 2013 INEQUALITIES 2 BMI and You For this assignment I will be demonstrating how you can find out how to use Inequalities to find whether or not you are overweight. You can use my equations and the examples of my work to find out exactly what category you might fall in. In this equation we are asked to use our own height to find out exactly if I might have a longer life span than the usual person. Before we get started, remember that W=my weight, and H=my height. W=176 and then H=65” In this first compound inequality, I had to take the equivalent inequality and replace it with
The area of a circle is approximately 3.14 times the radius squared. Which of the following expressions is a correct way to write this, if the radius is r? a. r(3.14)2 b. 3.14r2 c. (3.14r)2 d. 2(3.14)r 3. Merrill bought m notebooks for $2.50 each and n pens for $1.25 each.
To change kilograms to pounds [multiply (x) 2.2] 1kg = 2.2 pounds To change pounds into stones [divide (:-) 14] 1 stone = 14 pounds Weight Record Clients: Frank K start weight 48.90 kg Date Weight (kg) Loss/gain 06/01/13 48. 90 kg 0 13/01/13 48. 50 kg
b. One inch on this map equals exactly __24,000___ inches in real life. c. One inch on this map equals exactly _0.3__ mile(s) in real life, so we can say that 1 inch on the map equals approximately 1/3 mi on the ground. (Lab Man) 3 6. Review how the location of point x in Fig.9.5 in 9th ed’n was determined using PLS shorthand (see caption for Figure 9.5). Then, determine the location of point V in the Figure below using PLS shorthand.
Both of these formulas were found on page 225 in Mathematics in Our World (Bluman, 2005). Problem #37 • This sequence is geometric • Ending balance is $814.45 STEPS/CALCULTATIONS YOU PERFORMED TO REACH THE ANSWER: To find the ending balance, the formula of An = a1(rn-1) will be used. The initial balance is $500, the interest is 5%, and the time span is 10 years. 5% will be listed as 1.05 as the initial balance is 100% plus 5% interest, so 105% is written 1.05. The number of terms is n=10, the first term is a1=525, the common ratio is r = 1.05.
I had Emily carry eighty pounds of 20 lb. weights. I asked Emily to guess the amount of weight she was carrying, and she accurately guessed eighty pounds. I then gave Emily a single 40 lb. weight and asked her again to estimate the amount of weight she was carrying.
24 14400 = 1000(a + 1) Multiplication is carried out. 14400 1000(a + 1) Divide both sides by 1000. 1000 1000 14.4= a + 1 14.4– 1 = a + 1 – 1 Subtract 1 from both sides to isolate a. 13.4= a We have solved for a. The correct dose of Tamiflu is 600mg for a thirteen-year-old
8. A Graph was created using the second table with ‘X’ values in the y-axis and ‘f’ values in the x-axis. Results Table 1.1: Raw Results Weight (grams) | Trial | Values of f (in Newtons [N]) | Value of X (in centimetres [cm]) | 50 | 123 | 0.5 0.5 0.5 | 11 1 | 100 | 123 | 1 0.8 1 | 2 2.1 2 | 150 | 123 | 1.5 1.45 1.5 | 3 3.1 3.1 | 200 | 123 | 2 2 1.9 | 4 4.1 3.9 | 250 | 123 | 2.4 2.4 2.3 | 54.8 5 | Table 1.2: Averaged Results Weight (in grams) | | Value of X (in centimetres [cm]) | 50 | 0.5 | 1 | 100 | 0.93 | 2.03 | 150 | 1.48 | 3.06 | 200 | 1.96 | 4 | 250 | 2.3 | 4.93 | Distance Stretched (X) in Centimetres (cm) Distance Stretched (X) in Centimetres (cm) Force Applied (f) in Newtons (N) Force Applied (f) in Newtons (N) Analysis of Results The above results