In general, the highest price p per unit of an item at which a manufacturer can sell N items is not constant but is, rather, a function of N. Suppose

Nannie Mack 2021-02-25 Answered
In general, the highest price p per unit of an item at which a manufacturer can sell N items is not constant but is, rather, a function of N. 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.
\(\begin{array}{|c|c|} \hline Number\ N & Price\ p\\ \hline 250 & 52.50\\ \hline300 & 52.00\\\hline 350 & 51.50\\ \hline 400 & 51.00\\ \hline \end{array}\)
(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.
\(\displaystyle{R}=\)

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Expert Answer

Nathaniel Kramer
Answered 2021-02-26 Author has 13518 answers

Slope of line joining the points \(\displaystyle{\left({250},{52.50}\right)}{\quad\text{and}\quad}{\left({300},{52.00}\right)}{i}{s}\ {m}=\frac{{{y}{2}-{y}{1}}}{{{x}{2}-{x}{1}}}=\frac{{{52.00}-{52.50}}}{{{300}-{250}}}=-{0.01}\)
Slope of line joining the points \(\displaystyle{\left({300},{52.00}\right)}{\quad\text{and}\quad}{\left({350},{1.505}\right)}{i}{s}\ {m}=\frac{{{y}{2}-{y}{1}}}{{{x}{2}-{x}{1}}}=\frac{{{51.50}-{52.00}}}{{{350}-{300}}}=-{0.01}\) and so on.
Since slope of line joining the points (250, 52.50) and (300, 52.00) is equal to the slope of line joining the points (300, 52.00) and (350, 1.505) , so there is a linear relation between p and N
Let \(\displaystyle{p}={m}{N}+{b}\)
where \(\displaystyle{m}=-{0.01}\)
Therefore,
\(\displaystyle{p}=-{0.01}{N}+{b}\)
Now, the point (250, 52.50) lie on the line, so
\(\displaystyle{52.50}=-{0.01}\times{250}+{b}\)
\(\displaystyle{52.50}=-{2.5}+{b}\)
\(\displaystyle{b}={55}\)
Thus, \(\displaystyle{p}=-{0.01}{N}+{55}\)
(b) 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 is determined as follows
\(\displaystyle{R}={p}{N}\)
\(\displaystyle{R}={\left(-{0.01}{N}+{55}\right)}{N}\)
\(\displaystyle{R}=-{0.01}{N}^{2}+{55}{N}\)
Finally answer:
\(\displaystyle{\left({a}\right)}{p}=-{0.01}{N}+{55}\)
(b) \(\displaystyle{R}=-{0.01}{N}^{2}+{55}{N}\)

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Relevant Questions

asked 2021-03-05
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?
asked 2021-03-11

Let's say the widget maker has developed the following table that shows the highest dollar price p. widget where you can sell N widgets. Number N Price p \(200 53.00\)
\(250 52.50\)
\(300 52.00\)
\(35051.50\)

(a) Find a formula for pin terms of N modeling the data in the table.

(b) Use a formula to express the total monthly revenue R, in dollars, of this manufacturer in month as a function of the number N of widgets produced in a month. \(R=\) Is Ra linear function of N?

(c) On the basis of the tables in this exercise and using cost, \(C= 35N + 900\), use a formula to express the monthly profit P, in dollars, of this manufacturer asa function of the number of widgets produced in a month \(p=\) ?

asked 2021-06-13
1. Who seems to have more variability in their shoe sizes, men or women?
a) Men
b) Women
c) Neither group show variability
d) Flag this Question
2. In general, why use the estimate of \(n-1\) rather than n in the computation of the standard deviation and variance?
a) The estimate n-1 is better because it is used for calculating the population variance and standard deviation
b) The estimate n-1 is never used to calculate the sample variance and standard deviation
c) \(n-1\) provides an unbiased estimate of the population and allows more variability when using a sample and gives a better mathematical estimate of the population
d) The estimate n-1 is better because it is use for calculation of both the population and sample variance as well as standard deviation.
\(\begin{array}{|c|c|}\hline \text{Shoe Size (in cm)} & \text{Gender (M of F)} \\ \hline 25.7 & M \\ \hline 25.4 & F \\ \hline 23.8 & F \\ \hline 25.4 & F \\ \hline 26.7 & M \\ \hline 23.8 & F \\ \hline 25.4 & F \\ \hline 25.4 & F \\ \hline 25.7 & M \\ \hline 25.7 & F \\ \hline 23.5 & F \\ \hline 23.1 & F \\ \hline 26 & M \\ \hline 23.5 & F \\ \hline 26.7 & F \\ \hline 26 & M \\ \hline 23.1 & F \\ \hline 25.1 & F \\ \hline 27 & M \\ \hline 25.4 & F \\ \hline 23.5 & F \\ \hline 23.8 & F \\ \hline 27 & M \\ \hline 25.7 & F \\ \hline \end{array}\)
\(\begin{array}{|c|c|}\hline \text{Shoe Size (in cm)} & \text{Gender (M of F)} \\ \hline 27.6 & M \\ \hline 26.9 & F \\ \hline 26 & F \\ \hline 28.4 & M \\ \hline 23.5 & F \\ \hline 27 & F \\ \hline 25.1 & F \\ \hline 28.4 & M \\ \hline 23.1 & F \\ \hline 23.8 & F \\ \hline 26 & F \\ \hline 25.4 & M \\ \hline 23.8 & F \\ \hline 24.8 & M \\ \hline 25.1 & F \\ \hline 24.8 & F \\ \hline 26 & M \\ \hline 25.4 & F \\ \hline 26 & M \\ \hline 27 & M \\ \hline 25.7 & F \\ \hline 27 & M \\ \hline 23.5 & F \\ \hline 29 & F \\ \hline \end{array}\)
asked 2021-03-11
An automobile tire manufacturer collected the data in the table relating tire pressure x​ (in pounds per square​ inch) and mileage​ (in thousands of​ miles). A mathematical model for the data is given by
\(\displaystyle​ f{{\left({x}\right)}}=-{0.554}{x}^{2}+{35.5}{x}-{514}.\)
\(\begin{array}{|c|c|} \hline x & Mileage \\ \hline 28 & 45 \\ \hline 30 & 51\\ \hline 32 & 56\\ \hline 34 & 50\\ \hline 36 & 46\\ \hline \end{array}\)
​(A) Complete the table below.
\(\begin{array}{|c|c|} \hline x & Mileage & f(x) \\ \hline 28 & 45 \\ \hline 30 & 51\\ \hline 32 & 56\\ \hline 34 & 50\\ \hline 36 & 46\\ \hline \end{array}\)
​(Round to one decimal place as​ needed.)
\(A. 20602060xf(x)\)
A coordinate system has a horizontal x-axis labeled from 20 to 60 in increments of 2 and a vertical y-axis labeled from 20 to 60 in increments of 2. Data points are plotted at (28,45), (30,51), (32,56), (34,50), and (36,46). A parabola opens downward and passes through the points (28,45.7), (30,52.4), (32,54.7), (34,52.6), and (36,46.0). All points are approximate.
\(B. 20602060xf(x)\)
Acoordinate system has a horizontal x-axis labeled from 20 to 60 in increments of 2 and a vertical y-axis labeled from 20 to 60 in increments of 2.
Data points are plotted at (43,30), (45,36), (47,41), (49,35), and (51,31). A parabola opens downward and passes through the points (43,30.7), (45,37.4), (47,39.7), (49,37.6), and (51,31). All points are approximate.
\(C. 20602060xf(x)\)
A coordinate system has a horizontal x-axis labeled from 20 to 60 in increments of 2 and a vertical y-axis labeled from 20 to 60 in increments of 2. Data points are plotted at (43,45), (45,51), (47,56), (49,50), and (51,46). A parabola opens downward and passes through the points (43,45.7), (45,52.4), (47,54.7), (49,52.6), and (51,46.0). All points are approximate.
\(D.20602060xf(x)\)
A coordinate system has a horizontal x-axis labeled from 20 to 60 in increments of 2 and a vertical y-axis labeled from 20 to 60 in increments of 2. Data points are plotted at (28,30), (30,36), (32,41), (34,35), and (36,31). A parabola opens downward and passes through the points (28,30.7), (30,37.4), (32,39.7), (34,37.6), and (36,31). All points are approximate.
​(C) Use the modeling function​ f(x) to estimate the mileage for a tire pressure of 29
\(\displaystyle​\frac{{{l}{b}{s}}}{{{s}{q}}}\in.\) and for 35
\(\displaystyle​\frac{{{l}{b}{s}}}{{{s}{q}}}\in.\)
The mileage for the tire pressure \(\displaystyle{29}\frac{{{l}{b}{s}}}{{{s}{q}}}\in.\) is
The mileage for the tire pressure \(\displaystyle{35}\frac{{{l}{b}{s}}}{{{s}{q}}}\) in. is
(Round to two decimal places as​ needed.)
(D) Write a brief description of the relationship between tire pressure and mileage.
A. As tire pressure​ increases, mileage decreases to a minimum at a certain tire​ pressure, then begins to increase.
B. As tire pressure​ increases, mileage decreases.
C. As tire pressure​ increases, mileage increases to a maximum at a certain tire​ pressure, then begins to decrease.
D. As tire pressure​ increases, mileage increases.
asked 2021-04-20
(1 pt) A new software company wants to start selling DVDs withtheir product. The manager notices that when the price for a DVD is19 dollars, the company sells 140 units per week. When the price is28 dollars, the number of DVDs sold decreases to 90 units per week.Answer the following questions:
A. Assume that the demand curve is linear. Find the demand, q, as afunction of price, p.
Answer: q=
B. Write the revenue function, as a function of price. Answer:R(p)=
C. Find the price that maximizes revenue. Hint: you may sketch thegraph of the revenue function. Round your answer to the closestdollar.
Answer:
D. Find the maximum revenue. Answer:
asked 2021-06-05
Use the following Normal Distribution table to calculate the area under the Normal Curve (Shaded area in the Figure) when \(Z=1.3\) and \(H=0.05\);
Assume that you do not have vales of the area beyond \(z=1.2\) in the table; i.e. you may need to use the extrapolation.
Check your calculated value and compare with the values in the table \([for\ z=1.3\ and\ H=0.05]\).
Calculate your percentage of error in the estimation.
How do I solve this problem using extrapolation?
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...