As Viet drives around and checks meters to document the amount of electricity used in homes he would be part of the energy distribution process since the electricity is distributed off a power line into the house and then within the house by the wiring to the lights and the electrical outlets for usage by the inhabitants.
Answer:
public static void main(String[] args) {
String ing[] = {"ten","fading","post","card","thunder","hinge","trailing","batting"};
for (String i: ing){
if (i.endsWith("ing")){
System.out.println(i);
}
}
}
Explanation:
The for-loop cycles through the entire list and the if-statement makes it so that the string is only printed if it ends with "ing"
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)
Answer:
(a)Applications Time Stamp Events
(b)S=0.5(W-
)+, where ≤W≤+8.
Explanation:
Some applications assume that clocks always advance, so they could timestamp events under this assumption.
In our case we have the wrong timed clock, say W and the hardware clock H which is supposed to advance at a perfect rate.
We proceed to construct a software clock such that after 8 seconds we can replace the wrong timed clock with the software clock in good conditions.
Let us denote the software clock with S.
Then, S=c(W-)+ where:
=The current Time(10:27:54) and;
c is to be found.
We already know that S=+4 when W= +8,
So:
S=c(W-)+
+4=c(+8-)+
4=8c
c=0.5
We obtain the formula
S=0.5(W-)+, where ≤W≤+8.
