The rate 1.5 conpared to dignel of subculture of the domiance of the nukber hope it helps
Answer:
377 choices
Step-by-step explanation:
From the above question, we are told that
A restaurant offers 6 choices of appetizer, 8 choices of main meal and 5 choices of dessert. A customer can choose to eat just one course, or two different courses, or all three courses.
Let us represent each choice by :
A = Appetizer = 6
M = Main meal = 8
D = Dessert = 5
a) The combination of the 3 choices together
AMD=6 × 8 × 5=240
b) AM= Appetizer and Main meal
= 6 × 8 = 48
c) AD= Appetizer and Dessert
= 6 × 5 = 30
d) MD = Main meal × Dessert
= 8 × 5 = 40
e) A,M,D (each alone)=
Appetizer + Main meal + Dessert
= 6 + 8 + 5
= 19
Assuming all choices are available, how many different possible meals does the restaurant offer?
This is calculated as:
AMD + AM + AD + MD + A,M,D
240 + 48 + 30 + 40 + 19
= 377 choices
The answer for your question is 2(x+6)=28
To calculate P(at least one valve opens), we can find 1 - p(no valves open).
This is equal to 1 - (0.05)^5, or 0.9999996875.
P(at least one valve fails to open) is equal to 1 - p(all valves open and none fail). This is 1 - (0.95)^5, or 0.2262190625.
Its 81 because 729 divided by 3 and divided by 3 again is 81
