The local diner offers a meal combination consisting of an appetizer, a soup, a main course, and a dessert. There are three appetizers, five soups, three main courses, and three 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

number of ways to chose a three course meal is 90

Step-by-step explanation:

Given the data in the question;

Number of appetizer = 3

Number of soups = 5

Number of main courses = 3

Number of desserts = 3

Now,

Number of ways that appetizer can be chosen = ³C₁ = 3!/(1!(3-1)!) = 3

Number of ways that soups can be chosen = ⁵C₁ = 5!/(1!(5-1)!) = 5

Number of ways that main courses can be chosen = ³C₁ = 3!/(1!(3-1)!) = 3

Number of ways that desserts can be chosen = ³C₁ = 3!/(1!(3-1)!) = 3

So,

First we chose an appetizer, soup and main course.

Number of ways will be;

⇒ 3 × 5 × 3 = 45

Next,  we chose a dessert, a soup & a main course.

Number of ways will be;

⇒ 3 × 5 × 3 = 45

Total number of ways to chose a three course meal

⇒ 45 + 45 = 90 ways

Therefore, number of ways to chose a three course meal is 90

step-by-step explanation:

always plug in x. in this case, x equals 4.

i'm going to first solve 4(2+3x)

1) 4(2+3(4)) use pemdas to solve.

i'm going to multiply whats in the parenthesis first, so 3 times 4=12.

2)4 (2+12) i'm then going to multiply 4 by both numbers in the parenthesis (because multiplying comes before adding)

4 times 2= 8.   4 times 12= 48

4) (8+48)= 56 (answer)

i'm then going to solve 8+12x and plug in 4 for x.

1) 8 +12(4)

multiplying comes before adding, so i'm gonna multiply 12 and 4.

12 times 4= 48.

2) 8+48=56

both equal the same!

answer:

c 60% on edg