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Question # Suppose the manufacturer of widgets has developed the following table showing the highest price p, in dollars, of a widget at which N widgets can be s

Modeling data distributions
ANSWERED Suppose the manufacturer of widgets has developed the following table showing the highest price p, in dollars, of a widget at which N widgets can be sold.
$$\begin{array}{|c|c|} \hline Number\ N & Price\ p\\ \hline 200 & 53.00\\ \hline 250 & 52.50\\\hline 300 & 52.00\\ \hline 350 & 51.50\\ \hline \end{array}$$
(a) Find a formula for p in terms of N modeling the data in the table.
$$\displaystyle{p}=$$
(b) Use a formula to express the total monthly revenue R, in dollars, of this manufacturer in a month as a function of the number N of widgets produced in a month.
$$\displaystyle{R}=$$
Is R a linear function of N?
(c) On the basis of the tables in this exercise and using cost, $$\displaystyle{C}={35}{N}+{900}$$, use a formula to express the monthly profit P, in dollars, of this manufacturer as a function of the number of widgets produced in a month.
$$\displaystyle{P}=$$
(d) Is P a linear function of N? 2021-03-06
As per bartleby guidelines for the more than 3 subparts only first three are to be answered please upload the others separately.
(a)
Consider the two points from table (a) as (200 53.00) and (250 52.50).
Then, the equation is,
$$\displaystyle{p}-{p}_{{1}}=\frac{{{p}_{{2}}-{p}_{{1}}}}{{{N}_{{2}}-{N}_{{1}}}}{\left({N}-{N}_{{1}}\right)}$$
$$\displaystyle{p}-{53.00}=\frac{{{52.50}-{53.00}}}{{{250}-{200}}}{\left({N}-{200}\right)}$$
$$\displaystyle{p}-{53}=\frac{{-{0.50}}}{{50}}{\left({N}-{200}\right)}$$
$$\displaystyle{p}-{53}=-{0.01}{N}+{2}$$
$$\displaystyle{p}=-{0.01}{N}+{55}$$
Thus, the equation $$\displaystyle{p}=-{0.01}{N}+{55}$$
(b)
Consider the formula to express the total monthly revenue R, in dollars, of this manufacturer in a month as a function of the number N of widgets produced in a month.
$$\displaystyle{R}={\left(-{0.01}{N}+{55}\right)}{N}$$
$$\displaystyle=-{0.01}{N}^{2}+{55}{N}$$
Thus, the formula is $$\displaystyle{R}=-{0.01}{N}^{2}+{55}{N}.$$
Here, R is not the linear function of N.
(c)
The formula for the monthly profit is,
$$\displaystyle{P}=-{0.01}{N}^{2}+{55}{N}-{35}{N}-{900}$$
$$\displaystyle=-{0.01}{N}^{2}+{20}{N}-{900}$$
Here, P is not the linear function of N.