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    Mat 222 Week 1 Assignment

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    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

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    Mat 126 Week 1 Assignment

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    In this assignment I would like to talk about arithmetic sequences and geometric sequences and want to give an example each how to calculate with those sequences. First I want to give a short definition of each sequence. “An arithmetic sequence is a sequence of numbers in which each succeeding term differs from the preceding term by the same amount. This amount is known as the common difference.” (Bluman‚ A. G. 2500‚ page 221) An example for an arithmetic sequence is: 1‚ 3‚ 5‚ 7‚ 9‚ 11

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    Mat 540 Week 1 Assignment

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    Assignment Quantitative Methods Strayer University Step 1 For step‚ one I generated a column of random numbers‚ which will help me to determine the weeks between breakdowns. I used the equation in excel =Rand( ). Next‚ I generated a column to determine the weeks between breakdowns. I used the formula in excel =8*SQRT(r) and used the previous mentioned randomly generated numbers. Once I had the randomly generated weeks‚ I created a column for the cumulative

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    MATH 221 ASsignment 1

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    useful for simplifying algebraic expression because a lot of thing we do in life are equations. We use a lot of mathematical terms in the real world. Lastly I will show every step I took to simplify and identify each property of real numbers. The properties of algebra are important to know and understand. “Algebra is useful because it can be used to solve problems. Since problems are often communicated verbally‚ we must be able to translate verbal expressions into algebraic expressions and translate

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    MAT 222: Intermediate Algebra Title Page Solving a proportion as we learned this week‚ means that you are missing an import number in your equation or fraction‚ and you need to solve for that missing value. As in my example‚ I did not know what percentage of bills we each should pay. We knew each other’s salaries; but we really had to sit down‚ crunch numbers‚ and figure it out. For this week’s assignment we were asked to work through two proportions. For the first proportion‚ number

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    Mat 540-Assignment 1

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    days is random. It is best to generate this random‚ which is illustrated on the excel spreadsheet as r2. R2 will be numbers between 0 and 1. To determine this value a random number will be generated in excel. The number of days needed to repair the copier will be determined based on the cumulative outline below: 0.00 > r2 < 0.20‚ then it will take 1 day; 0.20 > r2 < 0.65‚ then it will take 2 days; 0.65 > r2 < 0.90‚ then it will take 3 days; and 0.90 > r2 < 1.00

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    Mat 221 Wk 5

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    then plugged the binomials into the Pyth. Thrm. Next I moved (2+6)^2 to le left of the equation by subtracting (2x+6)^2 from both sides. I then squared the expression Next I foiled the expression by multiplying and combining like terms. I multiplied -1 by each term inside the parentheses and then removed the parentheses around the expression (4x^2 + 16x + 16) Because x^2 an 4^2 are like terms I added 4x^2 to x^2 to get 5x^2 Since 5x^2 and -4x^2 are like terms I added -4x^2 to 5x^2 to get x^2

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    Mat 221 Cowling’s Rule

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    will have three variables. The formula used is d = D (a + 1). The following is the variables for the literal equation: a = child’s age – 5 Years old D = adult dose – 75 mg d = child’s dose I have been assigned to calculate a 5-year-old child’s dose of tamiflu given that the adult dose is 75mg. d = D (a + 1) The Cowling’s Rule formula 24 d = 75 (5 + 1) Substituted 75 for D and 5 for a. 24 d = 75 (6) Add 5+1 inside parenthesis‚ equals 6. 24 d = 450 Multiply

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    INTEGERS above are needed to solve each given expressions. A) A^3 – B^3 (7^3) – (-2^3) 343-(-6859) =7‚202 This is the given expression with VARIABLES A and B and raised to the EXPONENTS of 3 on each of them. By substituting the integers in the variables and raising them to the 3rd power gives the answer of B) (a – b)(a2 + ab + b2) (7-(-19) (7^2+(7)(-19)+(-19^2) (26) (49-133) + 361 (26)(277) =7202 C) ( b-c )/ (2b- a) :-This expression uses all three variables: a‚ b‚ and c. C)

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    This work MAT 126 Week 1 Assignment - Geometric and Arithmetic Sequence shows "Survey of Mathematical Methods" and contains solutions on the following problems: First Problem: question 35 page 230 Second Problem: question 37 page 230 Mathematics - General Mathematics Week One Written Assignment Following completion of your readings‚ complete exercises 35 and 37 in the “Real World Applications” section on page 280 of Mathematics in Our World . For each exercise‚

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