I have 22 questions

User Generated

fgnezbb

Mathematics

Description

Hello everyone, I have 22 questions for discrete mathematics course and I need a good expert who can helps me to solve them and its easy to solve them.

Unformatted Attachment Preview

Solve each of the problems below. To earn ANY CREDIT for each question you must SHOW YOUR WORK using clear and concise logic and proper mathematical notation. You’re graded on BOTH the accuracy of your answers AND your explanations that sufficiently support your answers. Unless otherwise stated, you’re to give EXACT VALUES of answers instead of decimal approximations. Regarding all applied problems: to earn ANY credit for such questions, you MUST declare a variable (unless the variable(s) has already been declared in the problem and set up and solve an appropriate mathematical expression that can be used to answer the question. Proper units must be included in your last answer. NO CREDIT WILL BE GIVEN FOR EITHER GUESSING OR CHECKING POSSIBLE ANSWERS WITHOUT SOLVING THE PROBLEM. Finally, you may write only your final answers on these pages; you must SHOW YOUR WORK both according to Homework Submission Guidelines and on your own paper. In order for this assignment to be graded, you must register on CONNECT. Courtesy access is acceptable. 1. In a few sentences, please explain why you’re taking this course in Discrete Mathematics. Dr. Farmer poses this question because he wants to insure proper service to the diverse group of majors represented in the course this semester. 2. Algebraically, find the point(s) of intersection between the line and the parabola. 3. A college charters a bus for $1700 to take a group to the museum. When six more students join the trip, the cost per student drops by $7.50. How many students were in the original group? In the new group, what is the cost per student? 4. The perimeter of a plot of land in the shape of a right triangle is 12 miles. If one leg of the triangle exceeds the other by 1 mile, find the length of each boundary of the land. 5. Suppose that we agree to pay you 8 cents for each problem in an exercise set that you solve correctly and fine you 5 cents for every problem done incorrectly. If at the end of 26 problems we do not owe each other money, how many problems did you solve correctly? 6. Parts of an automobile repair cost $175. The mechanic charges $34 per hour. If you receive an estimate for at least $226 and at most $294 for fixing the car, what is the time interval that the mechanic will be working on the job? 7. A baseball team plays in a large stadium. With a ticket price of $20, the average attendance at recent games has been 30,000. A market survey indicates that for each $1 increase in the ticket price, attendance decreases by 500. a. Express the number of spectators at a football game, N, as a function of the ticket price, x. b. Express the revenue from a football game, R, as a function of the ticket price, x. 8. A contractor is to build a warehouse whose rectangular floor will have an area of 4000 square feet. The warehouse will be divided into two rectangular rooms by an interior wall. The cost of the exterior walls is $175 per linear foot and the cost of the interior wall is $125 per linear foot. Express the contractor’s cost for building the walls, C, as a function of one of the dimensions of the warehouse’s rectangular floor, x. 9. An outdoor running track with semicircular ends and parallel sides is constructed. The length of each straight portion of the sides is 180 meters. The track has a uniform width of 4 meters throughout. Write the function that represents the area enclosed within the outer edge of the running track. 10. Refer to Problem #9. Write the function that represents the area enclosed within the inner edge of the running track. 11. Refer to Problems #9-10. What does the function represent? 12. Refer to Problems #9-11 as needed. Find the area of the track and the outer perimeter of the track. Round your answers to two decimal places. 13. A car rental agency can rent every one of its 200 cars at $30 per day. For each $1 increase in rate, five fewer cars are rented. Find the rental amount that will maximize the agency’s daily revenue. What is the maximum daily revenue? 14. Use the formula for compound interest with n compounding’s per year to solve this problem. How long, to the nearest tenth of a year, will it take $12,500 to grow to $20,000 at 6.5% annual interest compounded quarterly? 15. Use the formula for continuous compound interest to solve this problem. How long, to the nearest tenth of a year, will it take $50,000 to triple in value at 7.5% annual interest compounded continuously? 16. Does the parabola have a tangent whose slope is -1? If so, find an equation for the line (in slope-intercept form) for the line and the point of tangency. If not, why not? 17. Suppose the 100 largest cities in the United States in 1920 are ranked according to magnitude (areas of cities). From Lotka,1 the following relation holds approximately: Here, P is the population of the city having respective rank R. This relation is called the law of urban concentration for 1920. Solve for P in terms of R, and then find how fast the population is changing with respect to rank. 18. A 216 m2 rectangular pea patch is to be enclosed by a fence and divided into two equal parts by another fence parallel to one of the sides. What dimensions for the outer rectangle will require the smallest total length of fence? How much fence will be needed? 19. A manufacturer has determined that, for his product, the daily average cost (in hundreds of dollars) is given by ̅ √ As daily production increases, the average cost approaches a constant dollar amount. What is this amount? 20. Refer to problem #16. Determine the manufacturer’s marginal cost when 17 units are produced per day. 21. Refer to problem #16. The manufacturer determines that if production (and sales) were increased to 18 units per day, revenue would increase by $275. Should this move be made? Why? 22. Algebraically, find three ordered pairs such that √ when the solution is. (Note: there are many correct answers to this question).
Purchase answer to see full attachment
User generated content is uploaded by users for the purposes of learning and should be used following Studypool's honor code & terms of service.

Explanation & Answer

Please find attached the solution of all problems in word and PDF format. Please ask me if you have any doubt.Please note that Q-13 solution has been slightly changed from last solution. (Final answer is same, I have elaborated solution little bit more). Attached files have updated solution for that problem.Thanks.

1.

You have agreed to answer this.

2.

Equation of line: 𝑥 − 𝑦 = 2
Equation of parabola: 2𝑦 = 𝑥 2
To find intersection point:
From Equation of line, 𝑥 = 𝑦 + 2
Putting value of 𝑥 in Equation of parabola,
2𝑦 = (𝑦 + 2)2
⇒ 2𝑦 = 𝑦 2 + 4𝑦 + 4
⇒ 𝑦 2 + 2𝑦 + 4 = 0
⇒ (𝑦 2 + 2𝑦 + 1) + 3 = 0
⇒ (𝑦 + 1)2 = −3
This is not possible for any value of 𝑦, because (𝑦 + 1)2 ≥ 0∀𝑦 ∈ 𝑅
So, there is no point of intersection point between 𝒙 − 𝒚 = 𝟐 and 𝟐𝒚 = 𝒙𝟐 .

3.

Let, there were x students in the original group,
As, total cost = $1700
So, cost per student in original group =

total cost
number of students

=

$1700
x

As, 6 more students join the trip, number of students in new group = x + 6
As, for new group, cost per student drops by $7.50, so
1700
cost per student in new group =
− 7.5
x
As, for new group also, total cost should be $1700,
So,
cost per student in new group ∗ number of students in new group = 1700
1700
⇒(
− 7.5)(x + 6) = 1700
x
(1700 − 7.5x)(x + 6)

= 1700
x
⇒ (1700 − 7.5x)(x + 6) = 1700x
⇒ 1700x + 10200 − 7.5x 2 − 45x = 1700x
⇒ −7.5x 2 − 45x + 10200 = 1700x − 1700x
⇒ −7.5x 2 − 45x + 10200 = 0
⇒ −7.5(x 2 + 6x − 1360) = 0
⇒ x 2 + 6x − 1360 = 0
(Dividing both sides by -7.5)
⇒ x 2 + 40x − 34x − 1360 = 0
⇒ x(x + 40) − 34(x + 40) = 0
⇒ (x − 34)(x + 40) = 0
⇒ x = 34 or x = −40
As, number of students in the original group can’t be negative, so x = 34
So, there were 𝟑𝟒 students in the original group.
1700
1700
− 7.5 =
− 7.5 = 50 − 7.5 = 42.5
x
34
so, cost per student in new group is $ 𝟒𝟐. 𝟓
cost per student in new group =

4.

Let a, b and c are the short leg, long leg and hypotenuse of triangle respectively.
Here, 𝑏 = 𝑎 + 1
… (1)
also, perimeter = 12
⇒ 𝑎 + 𝑏 + 𝑐 = 12
⇒ 𝑐 = 12 − (𝑎 + 𝑏)
Putting 𝑏 = 𝑎 + 1 in above equation gives:
𝑐 = 12 − (𝑎 + 𝑎 + 1) = 12 − 2𝑎 − 1
⇒ 𝑐 = 11 − 2𝑎 … (2)
From Pythagoras Theorem:
𝑐 2 = 𝑎2 + 𝑏 2
Substituting 𝑐 and 𝑏 values from equation (1) and (2),
(11 − 2𝑎)2 = 𝑎2 + (𝑎 + 1)2
⇒ 121 − 44𝑎 + 4𝑎2 = 𝑎2 + 𝑎2 + 2𝑎 + 1
⇒ 121 − 44𝑎 + 4𝑎2 − 𝑎2 − 𝑎2 − 2𝑎 − 1 = 0
⇒ 2𝑎2 − 46𝑎 + 120 = 0
⇒ 2(𝑎2 − 23𝑎 + 60) = 0
⇒ 𝑎2 − 23𝑎 + 60 = 0(Dividing both sides by 2)
⇒ 𝑎2 − 3𝑎 − 20𝑎 + 60 = 0
⇒ 𝑎(𝑎 − 3) − 20(𝑎 − 3) = 0
⇒ (𝑎 − 20)(𝑎 − 3) = 0
⇒ a = 20 or a = 3
a = 20 is not possible, as side of a triangle can’t be greater than its perimeter.
So,
a=3
from, equation (1): 𝑏 = 𝑎 + 1 ⇒ b = 3 + 1 = 4
from, equation (2): 𝑐 = 11 − 2𝑎 ⇒ c = 11 − 6 = 5
so, length of each boundary of the land is: 3 miles, 4 miles and 5 miles.

5.

Let, I solved x problems correctly,
So, amount paid to me for correct solution = 8x
As, total problems solved are 26
So, number of problems solved incorrectly = (26 − x)
So, amount fined for incorrect solution = (26 − x) ∗ 5
As, we do not owe each other money
So, amount fined for incorrect solution = amount paid to me for correct solution
⇒ (26 − x) ∗ 5 = 8x
⇒ 130 − 5x = 8x
⇒ 130 = 13x
130
⇒x=
= 10
13
So, I solved 𝟏𝟎 problems correctly.

6.

If mechanic will be working on the job for x hours,
so mechanic charges for x hours = $34x
so, total cost = part cost + mechanic charges = 175 + 34x
minimum cost = $226
so, if xmin is minimum number of hours,
minimum cost = 175 + 34xmin
⇒ 226 = 175 + 34xmin
⇒ 226 − 175 = 34xmin
⇒ 51 = 34xmin
51

= xmin
34
⇒ xmin = 1.5 hours
maximum cost = $294
so, if xmax is maximum number of hours,
maximum cost = 175 + 34xmax
⇒ 294 = 175 + 34xmax
⇒ 294 − 175 = 34xmax
⇒ 119 = 34xmax
119

= xmax
34
⇒ xmax = 3.5 hours
So, the mechanic will be working on the job for 𝟏. 𝟓 − 𝟑. 𝟓 𝐡𝐨𝐮𝐫𝐬.

7.

As for each $1 increase in the ticket price, attendance decreases by 500
So, if there is $y increase in the ticket price,
Attendance will decrease by 500y.
a) ticket price after $y increase = present ticket price + y = 20+y
⇒ x = 20 + y
… (1)
Attendance after $y increase in the ticket price = present attendance – 500y = 30000 − 500y
⇒ N = 30000 − 500y
… (2)
From equation (1):
y = x − 20
Putting y’s value in equation (2):
N = 30000 − 500(x − 20)
⇒ N = 30000 − 500x + 10000
⇒ N = 40000 − 500x
So, the number of spectators at a football game, N, as a function of the ticket price, x:
𝐍 = 𝟒𝟎𝟎𝟎𝟎 − 𝟓𝟎𝟎𝐱
b) Revenue = ticket price*attendance = N*x
⇒ R = (40000 − 500x)x
⇒ R = 40000x − 500x 2
So, revenue from a football game, R, as a function of the ticket price, x:
𝐑 = 𝟒𝟎𝟎𝟎𝟎𝐱 − 𝟓𝟎𝟎𝐱 𝟐

8.

Let, x is the length of warehouse’s rectangular floor and y is the width of warehouse’s rectangular floor.
Area = Lenth ∗ width
⇒ xy = 4000
4000
⇒y=
… (1)
x
interior wall length = x
so, cost of interior wall = 125x
exterior wall length = perimeter of rectangular floor = 2(x + y)
so, cost of exterior wall = 175 ∗ 2(x + y) = 350(x + y)
so, contractor’s cost for building the walls = cost of interior wall + cost of exterior wall
⇒ C = 125x + 350(x + y)
Putting y =

4000
x

in above equation:

C = 125x + 350(x +

4000
x

)
1400000
x
𝟏𝟒𝟎𝟎𝟎𝟎𝟎
⇒ 𝐂 = 𝟒𝟕𝟓𝐱 +
𝐱
⇒ C = 125x + 350x +

9.

Let, the radius of inner semicircular ends is 𝑥 meters, so the radius of outer semicircular ends is 𝑥 + 4
meters.
So, length and width of outer rectangular portion are respectively: 180 meters and 2(𝑥+4) meters
So, area of outer recta...


Anonymous
Goes above and beyond expectations!

Studypool
4.7
Trustpilot
4.5
Sitejabber
4.4

Related Tags