The local diner offers a meal combination consisting of an appetizer, a soup, a main course, and a dessert. There are six appetizers, two soups, four main courses, and six desserts. Your diet restricts you to choosing between a dessert and an appetizer. (You cannot have both.) Given this restriction, how many three-course meals are possible?

96

Step-by-step explanation:

According to the fundamental principle of counting:

Total number of meals possible = (Choices for appetizers) x (Choices for soups) x (Choices for main courses) x (Choices for desserts)

If there is a restriction to choose between appetizer and dessert.

Then, Total number of meals possible = (Choices for appetizers) x (Choices for soups) x (Choices for main courses) +  (Choices for soups) x (Choices for main courses) x (Choices for desserts)

= 6 x 2 x 4 + 2 x 4 X 6

= 48+48

= 96

Hence, the number of three-course meals are possible= 96

3< x< 5

step-by-step explanation:

|x - 4| < 1

there are two solutions, one positive and one negative.   remember to flip the inequality when taking the negative solution

x-4 < 1       x-4 > -1

add 4 to each side

x-4+4 < 1+4       x-4+4 > -1+4

x< 5                 x> 3

3< x< 5

answer: c. is your answer

step-by-step explanation: get me a black leather coat