Answer:
Types of Relationships between the Input and Output
The scatter plot can be a useful tool in understanding the type of relationship that exist between the inputs (X’s) and the outputs (Y’s)
Step-by-step explanation:
1. No Relationship: The scatter plot can give an obvious suggestion if the inputs and outputs on the graph are not related. The points will be scattered throughout the graph with no particular pattern. For no relationship to exist, points have to be completely diffused. If some points are in concentration, then maybe a relationship does exist and our analysis has not been able to uncover it.
2. Linear and Non-Linear: A linear correlation exists when all the points are plotted close together. They form a distinct line. On the other hand points could be close together but they could form a relationship which has curves in it. The nature of the relationship has wide ranging implications.
3. Positive and Negative: A positive relationship between the inputs and the outputs is one wherein more of one input leads to more of an output. This is also known as a direct relationship.
On the other hand a negative relationship is one where more of one input leads to less of another output. This is also known as an inverse relationship.
4. Strong and Weak: The strength of the correlation is tested by how closely the data fits the shape. For instance if all the points are scattered very close together to form a very visible line then the relationship is strongly linear. On the other hand, if the relationship does not so obviously fit the shape then the relationship is weak.
I don't know if this was exactly what you were looking for; hope it is! :)
A quantity is said to be determined by exponential decay if it diminishes at a rate proportional to its value. The time essential for the decaying amount to descend to half of its initial value. Therefore the answer<span> is False.</span>
You multiply .65 times 100 and you get 65 which is 65% of 100
Don't listen to them asa is a very real thing! It's the angle side angle!! It would be C! Ssa! That doesn't exist!
<h2>
Answer with explanation:</h2>
It is given that:
f: R → R is a continuous function such that:
∀ x,y ∈ R
Now, let us assume f(1)=k
Also,
( Since,
f(0)=f(0+0)
i.e.
f(0)=f(0)+f(0)
By using property (1)
Also,
f(0)=2f(0)
i.e.
2f(0)-f(0)=0
i.e.
f(0)=0 )
Also,
i.e.
f(2)=f(1)+f(1) ( By using property (1) )
i.e.
f(2)=2f(1)
i.e.
f(2)=2k
f(m)=f(1+1+1+...+1)
i.e.
f(m)=f(1)+f(1)+f(1)+.......+f(1) (m times)
i.e.
f(m)=mf(1)
i.e.
f(m)=mk
Now,
Also,
i.e. 
Then,
(
Now, as we know that:
Q is dense in R.
so Э x∈ Q' such that Э a seq 
belonging to Q such that:
)
Now, we know that: Q'=R
This means that:
Э α ∈ R
such that Э sequence 
such that:
and
( since belongs to Q )
Let f is continuous at x=α
This means that:
This means that:
This means that:
f(x)=kx for every x∈ R
