Question

Two companies modeled their profits for one year.

Answer 1

We know the equation for company A is P(t) = 1.8(1.4)ᵗ.

and what we know about company B is that their equation is linear, and we also have 3 months in a table of values, and we can simply use two points from there to get the equation of company B, let's do so using (3,5) and (7, 25),

$\bf \begin{array}{ccccccccc} &&x_1&&y_1&&x_2&&y_2\\ % (a,b) &&(~ 3 &,& 5~) % (c,d) &&(~ 7 &,& 25~) \end{array} \\\\\\ % slope = m slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{25-5}{7-3}\implies \cfrac{20}{4}\implies 5 \\\\\\ % point-slope intercept \stackrel{\textit{point-slope form}}{y- y_1= m(x- x_1)}\implies y-5=5(x-3) \\\\\\ y-5=5x-15\implies y=5x-10\implies p(t)=5t-10$

now, the end of the year will be at the twelfth month, namely t = 12, and the four month well we know that t = 4.

how did each one fare?

$\bf \stackrel{\textit{at the 4th month}}{\stackrel{\textit{company A}}{P(4)=6.91488}\qquad \qquad \stackrel{\textit{company B}}{p(4)=10}} \\\\\\ \stackrel{\textit{at the end of the year}}{\stackrel{\textit{company A}}{P(12)\approx 102.04904}\qquad \qquad \stackrel{\textit{company B}}{p(12)=70}}$

so, notice, on the fourth month B did better than A, but at the end of the year, A ended up better off than B.

Answer 2

Answer:

Option B is the correct answer.

Step-by-step explanation:

Two companies modeled their profits for one year.

(1) Company A used the function P (t) = $(1.8) (1.4)^{t}$ to represent its monthly profit in hundreds of dollars after time t.

(2) Company B used the data of the table to write a linear model to represent its monthly income.

Let the linear function is P (t) = mt + c

where m is the slope of the line and c is y - intercept.

As per data we find two points (3.5) and (4.10) lying on the line.

so slope of the line = $\frac{y-y'}{x-x'}=\frac{10-5}{4-3}=5$

and point (3.5) passes through the given line

so,  5 = 5 × 3 + c

      5 = 15 + c

      c = 5 - 15 = - 10

So function will be P (t) = 5t-10

Now at t = 4

Company (A) profit P(4) =  $1.8(1.4)^{4}$ = 6.91

Company (B) profit p (4) = 5 × 4 - 10 = 10

so at t = 4 profits of Company B are more than Company A.

At t = 12

Company A  profit P (12) =  $1.8(1.4)^{12}$  = 102.05

Company B  profit P (12) = ( 5×12 ) - 10 = 66 - 10 = 50

So profit of company A is more at the end of the year.

Option B is the answer.