Assignment 33

 Mathematical word analysis:LINE: From the Latin word linum (flaxen chord). In ancient Rome, a flaxen chord was very sturdy and used for measuring lengths.

NOTE: This does not relate directly to the assignment above. It relates to mathematics in general.

Math is a language. Read the problem and clearly define any unknowns that you introduce. Some examples follow...

Example 1: An item with a list price of \$2,300 is advertised to be sold at a discount of 15%. What is the discounted price?

Solution: If x is the discounted price, then premise ==> x = (\$2,300)(1 - .15) = \$2,300(.85) = \$1,955.

Example 2: An item sells for \$700 after is has been discounted 20%. What was the original price?

Solution: If x is the original prices, the premise ==> .8x = \$700 ==> x = \$700/.8 = \$875.

Example 3: Working along, Herkimer can complete a job in 8 hours. If his sister, Hortense, works with him, they can complete the job in 3 hours. How long would it take Hortense to do the job if she worked along?

Solution: Let x = the number of hours it would take Hortense to do the job if she worked alone. Premise ==> that Herkimer and Hortense would complete 1/3 of the job in one hour if they worked together, and that Herkimer would complete 1/8 of the job in one hour if he worked alone. Our executive decision ==> Hortense would complete 1/x of the job in one hour if she worked alone. Hence, the ISH stage is

1/8 + 1/x = 1/3.

Solving this equation, we have 24x(1/8 + 1/x) = 24x(1/3) ==> 3x + 24 = 8x ==> 5x = 24 ==> x = 4.8. Our conclusion is that Hortense, if she worked alone, could complete the job in 4 hours and 48 minutes.

Here's a nice little thought question: Suppose that an item will sell at a 15% discount and that there is a 6% sales tax. Would you prefer to (a) discount the price, then apply the sales tax, or (b) apply the sales tax first, and then take the discount? (Don't just take a wild guess. Use MATH POWER and figure this out. It's really quite simple.)

 Problem: Use your calculator to fit a quadratic model to the points (-1,-9), (0,-4) and (1,5). Solution (with communication): Putting the x values in L1 and the y values in L2, and then using QuadReg L1,L2 we get the following output: y = ax2+bx+ca =2b=7c=-4R2=1 The quadratic model is y = 2x2 + 7x - 2 Since R2 = 1, the model is a perfect fit. This is not surprising, since three non-colinear points determine a unique parabola. Problem: Use your calculator to fit a quadratic model to the points (-1,-9), (0,-4), (1,5) and (3,12). Solution (with communication): Putting the x values in L1 and the y values in L2, and then using QuadReg L1,L2 we get the following output: y = ax2+bx+ca =-.6136363636b= 6.686363636c=-2.327272727R2=.9816446912 The quadratic model is approximately y = -.614x2 + 6.686x - 2.327 Since R2 is close to 1, the parabola is a reasonably good fit.