The vet told Jake that his dog, Rocco, who weighed 55 pounds, needed to lose 10 pounds. Jake started walking Rocco every day and
2 answers:
You might be interested in
Answer:
Linear Model: y(x)=35,000+750x
Step-by-step explanation:
In principle <em>Linear Models </em>are employed when we want to define a relationship between an independent and a dependent variable. <em>Linear Models</em> are also known as<em> Functions </em>, where one variable is expressed as a function of <em>at least one</em> other variable. The most common example would be a dependent variable that is directly linked as a response to any changes of an independent variable
. The most common expression of such relationship would be
where
is the relationship factor between
and
and is also known as the <em>slope </em>of the function (representing a rate of change of that function).
In this question we are told that the population of Bay Village today is 35,000. So we know this is our constant variable, lets call it .
Next we know that every year the population is increased by 750 people. So we know that our variable and thus our independent variable is every year which we shall call and our fixed factor is the 750 increment which we will call
.
From that we can conclude that:
<em>Year 1 Population: 35,000</em>
<em>Year 2 Population: 35,000 + 750 = 35,750</em>
<em>Year 3 Population: 35,750 + 750 = 36,500 </em>
And so on and so forth.<em> </em>
So we want to express the above as a Linear Model of the form:
<em> </em>Eqn(1).
Thus from Eqn (1) and all information given (i.e. pluging in values) we finally obtain:
Which for every different value of (i.e. every year) we can obtain the new and increased population
.