Answer:
The answer is "5 users and 1 block".
Explanation:
In Option a:
Bandwidth total 
Any User Requirement 
The method for calculating the number of approved users also is:
Now, calculate the price of each person for overall bandwidth and demands,


In Option b:



mean user 

max user 

Answer:
Let P(x) = x is in the correct place
Let Q(x) = x is in the excellent place
R(x) denotes the tool
Explanation:
a) Something is not in the correct place.
P(x) is that x is in the correct place so negation of ¬P(x) will represent x is not in the correct place. ∃x is an existential quantifier used to represent "for some" and depicts something in the given statement. This statement can be translated into logical expression as follows:
∃x¬P(x)
b) All tools are in the correct place and are in excellent condition.
R(x) represents the tool, P(x) represents x is in correct place and Q(x) shows x is in excellent place. ∀ is used to show that "all" tools and ∧ is used here because tools are in correct place AND are in excellent condition so it depicts both P(x) and Q(x). This statement can be translated into logical expression as follows:
∀ x ( R(x) → (P(x) ∧ Q(x))
c) Everything is in the correct place and in excellent condition.
Here P(x) represents correct place and Q(x) represents excellent condition ∀ represent all and here everything. ∧ means that both the P(x) and Q(x) exist. This statement can be translated into logical expression as follows:
∀ x (P(x) ∧ Q(x)
A .jpg file is going to be a picture. =)
Answer:
b. structured
Explanation:
Based on the information being described within the question it can be said that the type of decisions being mentioned are known as structured decisions. These are decisions which have various processes in place in order to handle a certain situation. Usually due to the problem having occurred countless times and are predictable.