Solve each system of equation. y = 2x – 5 5. x + y = - 7 6.1 Warm-up Solve each equation. 4x – 10 = 8 2z + 3(4z + 5) = 20 5(2y + 1) = 4y + 12 Solve each system of equation. y = 2x – 5 5. x + y = - 7 3y – x = 5 3x + y = - 9 x = 9/2 = 4.5 z = 5/14 = .357 y = 7/6 = 1.167 (4, 3) (- 1, - 6)

Counting Principle

In situations where we consider combinations of items or a succession of events such as flips of a coin or the drawing of cards, each result is called an outcome. An event is a subset of all possible outcomes. When an event is composed of two or more outcomes, such as choosing a card followed by choosing another card, we have a compound event. Counting Principle: Suppose an event can occur in p different ways. Another event can occur in q different ways. There are pq ways both events can occur.

n! = n(n – 1)(n – 2)(n – 3)…(3)(2)(1) Find the number of possible outcomes. 1. Jani can choose gray or blue jeans, a navy, white, green, or striped shirt and running shoes, boots, or loafers. How many outfits can she form? 2 X 4 X 3 = 24 outfits

2. A restaurant offers four sizes of pizza, two types of crust, and eight toppings. How many combinations of pizza with one topping are there? 3. In how many different ways can Annie, Bob, Carmine, and Della be arranged in a line? 4 X 2 X 8 = 64 pizzas 4! 4!=4 X 3 X 2 X 1 =24

4. In how many ways can 6 people line up at a ticket window. 5 4. In how many ways can 6 people line up at a ticket window? 5. Evaluate 9! 6. Rewrite 12! with a factor of 10! 6! 6! = 6 x 5 x 4 x 3 x 2 x 1 = 720 9! = 9 x 8 x 7 x6 x 5 x 4 x 3 x 2 x 1 =362880 12! = 12 x 11 x 10!

Estimating the number of people in a large crowd (for example, watching a parade or attending/marching in a political rally) is quite challenging and often leads to controversies. One method sometimes used is to focus on a small section of the crowd, such as a rectangular area. 1. Make a square measuring 5 feet by 5 feet, and have your friends stand inside it as if they are watching a band at a small club. Count the number of your friends that comfortably fit in the rectangle and find the ratio of this number to the rectangle’s area. Explain in your own words what this ratio means. The area is 25 so the ratio is

2. Use this value to estimate the size of a crowd that is 10 feet deep on both sides of the street standing along a 1-mile section of a parade route. There are four 5-foot-by-5-foot squares in a 10-foot-by-10-foot section. There are 5,280 feet in 1 mile, so there are 528 10-foot-by-10-foot sections. To include both sides of the route, a factor of 2 is included. 2 • 4E • 528 = 4,224E, or somewhat more than 4,000E

3. One rule of thumb for estimating crowds is that each person occupies 2.5 square feet. Use this rule to estimate the size of the crowd watching a parade along the 1-mile section of the route in Question 2. A good mental (lower) estimate is 2 • 40 • 500 = 40,000 since there are 100/2.5 = 40 people in each 10-foot-by-10-foot section. For mental estimates, 5,000 feet can be used for the number of feet in 1 mile. So, there are about 500 10-foot-by-10-foot sections in a mile.

A telephone number in the form NYZ-ABC-XXXX has three sections: NYZ ABC XXXX area code exchange code station code Before 1995, all area codes had the form NYZ, where N was any digit from 2 to 9 The restrictions on N saved 0 for call operator and 1 for long-distance calls. In addition, codes such as 800 and 911 were (and still are) used for special purposes. The restriction that Y = 0 or 1 was removed in 1995 because all possible area codes had been assigned. Today N is 2–9, Y is 0–8, and Z is 0–9; the exception to these rules are codes of the form 37Z and 96Z, which are being reserved for future use. Area codes where Y = Z are called easily recognizable codes and are often assigned to special services such as 800 and 877.

How many area codes were possible before 1995? 2. According to the post-1995 rules, how many area codes are possible today? 8 • 1 • 8 + 8 • 1 • 9= 64 + 72 = 136 8 • 9 • 10 − 1 • 1 • 10- 1 • 1 • 10 = 700 There are, in fact, a few other restrictions that reduce the number of available area codes to 681. As of September 2008, 377 area codes have been assigned.

3. The 7-digit numbers in a given area code have the form ABC-XXXX, where X, B, and C can be any digit 0–9 and A is restricted to 2–9. There are two other restrictions: B and C cannot both equal 1 since these values are designated for other purposes such as 911 (emergency) and 411 (information), and 555-0100 through 555-0199 are reserved for fictional uses such as in television shows or movies. According to these conditions, how many 7-digit numbers are possible in a single area code? 8 • 106 − 8 • 1 • 1 • 104 − 100 [fictional numbers] = 7,919,900 Homework: page 648-649 (1-6,11,12,17-20,25,26)