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03-62-194 Practice Questions Last Update: March 21, 2013 Many of these exercises are from the course text book Mathematical Applications For the Management, Life and Social Sciences, Second Custom Edition for 03-62-194. This custom edition consists of selections from Mathematical Applications For the Management, Life and Social Sciences”, Eighth Edition, by Ronald J. Harshbarger and James J. Reynolds. The publisher is Nelson Education. Other questions my come from the For many of the questions solutions are not given to encourage you to try the problems yourself or with a group in your class. The odd numbered questions from the text have an answer in the back of the text so you can check your solution. I am happy to discuss any of the practice questions in the tutorial or during my oﬃce hours. You can also get help from the Math and Stats Learning Centre (ER 3125).

(1) You need to borrow \$100, 000 to start your business. You can borrow from the bank at 10% interest and you can borrow from your life insurance policy at 12%. Your business plan budgets for \$10, 100 in interest payments in the ﬁrst year. How much do you borrow from the bank, and how much do you borrow from your life insurance? Solution. Step 1: Deﬁne the variables. Let B be the amount, in dollars, to be borrowed from the bank and let L be the amount, in dollars, to be borrowed from the life insurance policy. Step 2: Set up the mathematical problem. We have two equations. The ﬁrst equation states that the total amount to be borrowed is \$100, 000. That is, we must have B + L = 100, 000. The second equation states that the interest to be paid is \$10, 100. That is, 0.1B + 0.12L = 10, 100. Together they give the following system of two equations in two variables. B + L = 100, 000 0.1B + 0.12L = 10, 100 (1) (2)

Step 3: Solve the mathematical problem. We create the augmented matrix and reduce it. We have 1 1 100, 000 . 0.1 0.12 10, 100 Complete the operation Row2 − 0.1 ∗ Row1 to get 1 1 100, 000 0 .02 100 1 1 100, 000 0 1 5, 000
1

.

Multiply Row2 by 50 to get .

Now do the operation Row1 − Row2 to get 1 0 95, 000 0 1 5, 000 .

We read the matrix to get B = 95, 000 (ﬁrst row) and L = 5, 000 (second row). Step 4: State the solution to the word problem. To follow the business plan borrow \$95, 000 from the bank and \$5, 000 from the life insurance policy. (2) (Section 1.1 # 53) A retired women has \$120, 000 to invest. She has chosen one relatively safe investment fund that has an annual yield of 9% and another, riskier one that has a 13% annual yield. How much should she invest in each fund if she would like to earn \$12, 000 in interest after one year? (3) A student must earn a grade of at least 93% to get a grade of A+ in a course. There are three tests each worth 20% of the ﬁnal grade. The ﬁnal exam is worth 40% of the ﬁnal grade. The student earned 92% on the ﬁrst test, 90% on the second test, and 91% on the third test. What is the lowest grade the student can score on the ﬁnal exam in order to get an A+ on the course? Solution. Let P to be the percentage mark on the ﬁnal exam. Now, how do we calculate the ﬁnal grade? On the ﬁrst test, the student earned 92% of the available points. There are twenty points available, so the student earned .92 × 20 = 18.4 marks. On the second test .90 × 20 = 18 marks were earned. On the third test .91 × 20 = 18.2 marks. There are 40 points available from the ﬁnal exam and the student will earn P × 40 of those points. The mathematical model is to determine the value of P so that 18.4 + 18 + 18.2 + 40P ≥ 93 ←→ 40P ≥ 38.4 ←→ P ≥ 0.96. In order to earn a grade of A+, the student must earn a mark of at least 96% on the ﬁnal exam. (4) A manufacturer sells its product for \$50 per unit. The ﬁxed costs are \$10, 000 and the variable costs are \$30 per unit. Let x be the number of units produced and sold. (a) (1 mark) Give an equation for the cost function C(x). (b) (1 mark) Give an equation for the revenue function R(x). (c) (1 mark) Give an equation for the proﬁt function P (x). (d) (2 marks) How many units must be sold to break-even? Solution. We have (a) C(x) = 30x + 10, 000. (b) R(x) = 50x. (c) P (x) = R(x) − C(x) = 20x − 10, 000. (d) Break-even is when P (x) = 0 which is when x = 500. (5) (Section 1.1 # 57) For a certain product, the revenue function is R(x) = 40x and the cost function is C(x) = 20x + 1600. To obtain a proﬁt, the revenue must be greater than the cost. For what values of x will there be a proﬁt?
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(6) (Section 1.1 # 61) The total amount spent by the USA for wireless communication services, S (in billions of dollars), can be modeled (that is, approximated with some accuracy) by S = 6.205 + 11.23t where t is the number of years past 1995. (Source: Cellular Telecommunications and Internet Association). (a) What value of t represents the year 2005? (b) What values of t give S > 150? (c) In what year does this equation project that spending for wireless communication services will exceed \$150 billion? (7) (Section 1.3 #56) Residential customers who heat their home with natural gas have their monthly bills calculated by adding a base service charge of \$5.19 and an energy charge of 51.91 cents per hundred cubic feet of natural gas that is used. Write an equation for the monthly charge y in dollars in terms of x, the number of hundreds of cubic feet used. (8) (Section 1.5 # 47) A concert promoter needs to take in \$380, 000 on the sale of 16, 000 tickets. If the promoter charges \$20.00 for some tickets and \$30.00 for others, how many of each type of ticket must be sold to yield the \$380, 000?

(9) (Section 1.3 #1) Find the intercepts and graph the line given by the equation 3x + 4y = 12. (10) (Section 1.3 #5) Find the intercepts and graph the line given by the equation 3x + 2y = 0. (11) (Section 1.3 #7) What is the slope of the line passing through the two points (22, 11) and (15, −17)? (12) (Section 1.3 #11) What is the slope of a horizontal line? What is the slope of a vertical line? (13) (Section 1.3 #17) What is the slope and intercept of the line given by y = (7/3)x − (1/4)? Graph the line. (14) (Section 1.3 #25) Graph and give an equation for the line with slope (1/2) and y−intercept −3. (15) (Section 1.3 #31) Graph and give an equation for the line with slope −(3/4) that passes through the point (−1, 4). (16) (Section 1.3 #39) Are the two lines given by the equations 3x + 2y = 6 and 2x − 3y = 6 parallel, perpendicular, or neither? (17) (Section 1.3 #43) Give an equation for the line through (−2, −7) that is parallel to the line given by 3x + 5y = 11.

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(18) The cost of a business property is \$1, 000, 000 and a company will use a straight-line depreciation schedule. It is estimated that the property will have a residual value of \$50, 000 after 25 years. If y is the value of this property after x years, then write the equation of this depreciation schedule. After how many years will the value of the property be \$12, 000. Solution. Set (x1, y1 ) = (0, 1, 000, 000) and (x2, y2 ) = (25, 50, 000). The slope is m= 50, 000 − 1, 000, 000 y2 − y1 = = −38, 000. x2 − x1 25 − 0

The equation is y = mx+b = −38, 000x+b. To get b we substitute (x1, y1 ) = (0, 1, 000, 000) into the equation to get 1, 000, 000 = −38, 000(0) + b ←→ b = 1, 000, 000. The equation of depreciation is y = −38, 000x + 1, 000, 000. We now solve 12, 000 = −38, 000x + 1, 000, 000 ←→ x = 26 and conclude that the value is \$12, 000 after 26 years. (19) (Section 1.3 #47) A \$360, 000 building is depreciated by its owner. The value y of the building afterx months of use is y = 360, 000 − 1, 500x dollars. (a) Graph the function for x ≥ 0. (b) How long is it until the building is completely depreciated? (c) The point (60, 270000) lies on the graph. Explain what this means. (20) (Section 1.1 # 41) A \$648, 000 property is depreciated for tax purposes by its owner with the straight-line depreciation method. The value of the building, y, after x months is given by y = 648, 000 − 1800x dollars. After how many months is the value of the building \$387, 000? (21) For each matrix given below, explain why it is not in reduced form.   1 0 0 1 (a) 0 1 0 1 0 0 6 1   1 0 0 1 (b) 0 1 0 1 0 1 1 1 (22) Solve the following system by reducing the augmented matrix. 3x − 4y = 1 −x + 2y = 1
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(3 marks)

(23) Solve the following system by reducing the augmented matrix. 8x − 10y = 100 −4x + 5y = −502 (24) Solve the following system by reducing the augmented matrix. 2x − 3y = −11 4x − 6y = −23 (25) Solve the following system by reducing the augmented matrix. 2x − y + z = 5 −3x + y + 2z = −1 x + 2y + 4z = 12 (26) Solve the following system by reducing the augmented matrix. x − y + z = 71 x − y + 2z = 12 2x − 2y + 4z = 24 (27) Solve the following system by reducing the augmented matrix. x1 x1 2x1 + x2 − x3 x2 + x3 − 2x3 − x2 − 5x3 + x4 + x4 + x4 − 3x4 = = = = 3 1 6 −5

(28) Solve the following by reducing the augmented matrix. x1 + 2x2 − x3 = 3 3x1 + x2 = 4 2x1 − x2 + x3 = 2

(29) (Section 3.3 # 13) Solve the following by reducing the augmented matrix. x + y − z = 0 x + 2y + 3z = −5 2x − y − 13z = 17

(30) (Section 3.3 # 15) Solve the following by reducing the augmented matrix. 2x − 6y − 12z = 6 3x − 10y − 20z = 5 2x − 17z = −4 x − 3y + 3z = 7 x + 2y − z = −2 3x + 2y + 4z = 5

(31) (Section 3.3 # 17) Solve the following by reducing the augmented matrix.

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(32) Solve the following by reducing the augmented matrix. x + 2y − z = 10 2x − 4y + 5z = 4 3x − 2y + 4z = 11 (33) Solve the following by reducing the augmented matrix. x + y − 2z = −3 2x − y + 3z = 7 x − 2y + 5z = 0 (34) Solve the following by reducing the augmented matrix. x − y + 2z = 1 2x + y − 3z = 1 5x − 2y + 3z = 6 (35) Solve the following by reducing the augmented matrix. x − 3y + 2z = −4 3x − 9y + 6z = −12 2x − 6y + 4z = 8 (36) Solve the following by reducing the augmented matrix. x + y − 2z = −3 2x − y + 3z = 7 x − 2y + 5z = 0 (37) Reduce the augmented matrix to echelon form then give the solution. x + 2y − 3z = 8 −2x + 2y + z = 3 3x − 2y + 8z = 9  1 2 −3 8 1 3  R2 + 2R1 [A|b] =  −2 2 3 −2 8 9 R3 − 3R1   1 2 −3 8  0 6 −5 19  R2/6 0 −8 17 −15   1 2 −3 8 R1 − 2R2  0 1 −5/6 19/6  0 −8 17 −15 R3 + 8R2 
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We have

We read oﬀ the solution as x = 3, y = 4, and z = 1.

 1 0 −4/3 5/3  0 1 −5/6 19/6  0 0 31/3 31/3 (3/31)R3   1 0 −4/3 5/3  0 1 −5/6 19/6  0 0 1 1   1 0 0 3  0 1 0 4  0 0 1 1

(38) Reduce the augmented matrix to echelon form then give the solution. x + 2y − 3z = −2 3x − y − 2z = 1 2x + 3y − 5z = −3 Solution. We have  1 2 −3 −2 [A|b] =  3 −1 −2 1  R2 − 3R1 2 3 −5 −3 R3 − 2R1   1 2 −3 −2  0 −7 7 7  −R2/7 1 0 −1 1   1 2 −3 −2 R1 − 2R2  0 1 −1 −1  1 R3 + R2 0 −1 1   1 0 −1 0  0 1 −1 −1  0 0 0 0 

The coeﬃcient matrix is in reduced echelon form and we read of the solution as x = z, y = −1 + z and z can be any real number. (39) State all nonnegative, integer solutions, to the system of linear equations with the following reduced augmented matrix.   1 0 −2 6  0 1 1 3 . 0 0 0 0

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(40) (Section 1.5 # 53) A social agency is charged with providing services to three types of clients, A, B, and C. A total of 500 clients are to be served, with \$150, 000 available for counseling and \$100, 000 available for emergency food and shelter. Type A clients require an average of \$200 for counseling and \$300 for emergencies. Type B clients require an average of \$500 for counseling and \$200 for emergencies. Type C clients require an average of \$300 for counseling and \$100 for emergencies. How many of each type of client can be served? Solution. Let A, B and C be the number of type A, B and C clients to be served, respectively. We get three equations. The ﬁrst is to ensure that the number served is 500 and the next two are to ensure we spend the total counseling and emergency (food and shelter) budgets. The system of equations is A + B + C = 500 200A + 500B + 300C = 150 000 300A + 200B + 100C = 100 000. The augmented matrix is We have   500 1 1 1 [A|b] =  200 500 300 150 000  R2 − 200R1 R3 − 300R1 300 200 100 100 000   1 1 1 500  0 300 50 000  R2/3 100 0 −100 −200 −50 000   1 1 1 500 R1 − R2  0  500/3 1 1/3 0 −100 −200 −50 000 R3 + 100R2   1 0 2/3 1000/3  0 1  1/3 500/3 R3 ∗ (−3/500) 0 0 −500/3 −100000/3   1 0 2/3 1000/3 R1 − (2/3) ∗ R3  0 1 1/3 500/3  R2 − (1/3) ∗ R3 0 0 1 200 1 0 0  0 1 0 0 0 1 Since the coeﬃcient matrix is in reduced   200 100  . 200 row echelon form we are done and can read the

solution as A = 200, B = 100 and C = 200. The social agency can serve 200 type A clients, 100 type B clients and 200 type C clients. (41) An adventure company has three types of vehicles that carry three types of cargo. The capacity of each type is summarized in the table below. Suppose that on a given weekend the adventure company must transport 31 passengers, 19 canoes and kayaks, and 370 kilograms of luggage. Set up, but do not solve, the system of linear equations whose
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solution will determine the number of each vehicle type to be scheduled. Be sure to include a well written deﬁnition of the variables. Vehicle Type Units Carried SUV Van Car Passengers 3 7 4 Canoes/Kayaks 3 4 1 Luggage (in kilograms) 100 50 20 Solution. Let S, V , and C be the number of SUVs, vans, and cars, respectively, to be scheduled. The following equations ensure that the number of passengers and canoes and kayaks; and the amount of luggage to be carried meets the requirements. 3S + 7V 3S + 4V 100S + 50V + 4C = 31 + C = 19 + 20C = 370.

(42) (This is similar to Section 3.4 # 64) The King Trucking Company has an order for three products for delivery. The following table gives the particulars (volume, weight, and value) of each product. Deﬁne the variables and set up, but do not solve the system of equations that would determine the number of each product that can be transported if the capacity is 6,000 cubic meters and 11,000 kilograms; and if their insurance policy can cover \$36,900 of transported goods. Product 1 Product 2 Product 3 Unit Volume (cubic metres) 10 8 20 10 20 40 Unit weight (kilograms) Unit values (dollars 100 20 200 (43) (Section 1.6 # 46) Find the market equilibrium point when the demand function is p = 480 − 3q and the supply function is p = 17q + 80. (44) (Section 1.6 # 49) A group of retailers will buy 80 televisions from a wholesaler if the price is \$350 and 120 if the price is \$300. The wholesaler is willing to supply 60 if the price is \$280 and 140 if the price is \$370. Assuming that the supply and demand functions are linear, ﬁnd the equilibrium point for the market. (45) Find the equilibrium point when the supply equation is p = 10q + 50 and the demand equation is p = 600 − q. Solution. We solve the two equations for p and q. Using the two equations we see that, since p = p, 10q + 50 = 600 − q ←→ q = 50. (2 marks) When q = 50, we have Market equilibrium is when p = 550 and q = 50.(1 marks) p = 10(50) + 50 = 550 or p = 600 − 50 = 550. (2 marks)

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(46) Find the inverse of the following matrix: Solution. We start with [A|I ] =

1 4 . 4 18

1 4 1 0 4 18 0 1

.

We take row 2 minus 4 times row 1 (R2 − 4R1) to get [A|I ] = Now divide row 2 by 2 to get [A|I ] = Now take R1 − 4R2 to get [A|I ] = 1 0 9 −2 0 1 −2 1/2 9 −2 −2 1/2 4 7 ? 1 2 2 −1 ? −4 2 . 1 4 1 0 0 1 −2 1/2 . 1 4 1 0 0 2 −4 1 .

Since A reduced to the identity, we know that A−1 = .

(47) (Section 3.4 # 5) What is the inverse of

(48) (Section 3.4 # 7) What is the inverse of

  3 0 0 (49) (Section 3.4 # 3) What is the inverse of 0 3 0 ? 0 0 3 (50) (Section 3.4 # 25) If A−1 = 3 2 3 . , Solve Ax = 1 1 2

(51) Find A−1

  1 1 0 when A = 1 0 2 . 0 −1 1

(52) Use the inverse matrix to solve 2x = 10 . x + y = 12
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(53) (Section 3.4 # 3) Does B  1 A = 0 1 (54) Use inverses to solve

= A−1 when    2 1 0 −1/3 1 0 3 and B = 1/2 0 −1/2? 0 1 0 1/3 0

x1 + x2 + x3 = 3 2x1 + x2 + x3 = 4 2x1 + 2x2 + x3 = 5 (55) Set up, but do not solve the LP for the following problem. Be sure to include a well written deﬁnition of the variables. A sausage company makes two kinds of wieners, regular and all-beef. Each kilogram of all-beef wieners requires 0.75 kg of beef and 0.2 kg of spices. Each kg of regular wieners requires 0.18 kg of beef, 0.3 kg of pork and 0.2 kg of spices. Suppliers can deliver at most 1020 kg of beef, at most 600 kg of pork, and at most 500 kg of spices. If proﬁt is \$0.60 on each kg of all-beef wieners and \$0.40 on each kg of regular wieners, how many kilograms of each should be produced to maximize proﬁt? (Text: Section 4.2 # 43.) (56) Set up, but do not solve the LP for the following problem. Be sure to include a well written deﬁnition of the variables. The Wellbuilt Company produces two types of wood chippers, economy and deluxe. The deluxe model requires 3 hours to assemble and 1/2 hour to paint; and the economy model requires 2 hours to assemble and 1 hour to paint. The maximum number of assembly hours available is 24 hours per day; and the maximum number of painting hours available is 8 hours per day. If the proﬁt on the deluxe model is \$15 per unit and the proﬁt on the economy model is \$12 per unit, How many of each unit should be produced each day to maximize proﬁt? (Text: Section 4.2 # 27.) (57) Set up, but do not solve the LP for the following problem. Be sure to include a well written deﬁnition of the variables. A company manufactures two types of electric hedge trimmers, one of which is cordless. The cord type trimmer requires 2 hours to make and the cordless type trimmer requires 4 hours. The company has only 800 work hours in manufacturing each day; and the packaging department can only package 300 trimmers per day. If the company sells the cord type for \$22.50 and the cordless for \$45.0, how many of each type should it produce per day to maximize its sales. (Text: Section 4.2 # 29.) (58) Set up, but do not solve the LP for the following problem. Be sure to include a well written deﬁnition of the variables. A ﬁrm manufactures bumper bolts and fender bolts for antique cars. One machine can produce up to 130 fender bolts per hour and the other machine can produce up to 120 bumper bolts per hour. The combined number of fender bolts and bumper bolts that can be packaged is 230 per hour. How many of each type of bolt should be produced each hour to maximize sales if fender bolts sell for \$1 and bumper bolts sell for \$2? (Text: Section 4.2 # 30.)
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(59) (Exercises 4.1: # 19) Graph the solution of the system x + 2y x + y 2x + y x y ≤ ≤ ≤ ≥ ≥ 48 30 50 0 0

(60) (Exercises 4.1: # 23) Graph the solution of the system x + 3y ≥ 3 2x + 3y ≥ 5 2x + y ≥ 3 x ≥ 0 y ≥ 0 (61) Graph the feasible region deﬁned by the constraints given below. Shade the feasible region and provide the coordinates of all the corner points. 2x + y y x y ≤ ≤ ≥ ≥ 6 3 0 0

(62) Graph the feasible region deﬁned by the constraints given below. Shade the feasible region and provide the coordinates of all the corner points. 3x + 2y y x y ≤ ≤ ≥ ≥ 12 3 0 0

(63) Draw the feasible region deﬁned by the constraints given below. Shade the feasible region and provide the coordinates of all the corner points. 6x + 2y y x y ≤ ≤ ≥ ≥ 24 3 0 0

(64) Draw the feasible region deﬁned by the constraints given below. Shade the feasible region and provide the coordinates of all the corner points. 2x + 2y y x y ≤ ≤ ≥ ≥ 8 3 0 0

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(65) Solve the LP using the graphical method. minimize subject to −x − y = z x + 2y ≤ 6 2x + y ≤ 6 x ≥ 0 y ≥ 0

(66) Solve the LP using the graphical method. minimize subject to

(67) Solve the LP using the graphical method. minimize subject to

−x − y = z 3x + 2y ≤ 12 x − y ≤ 1 x ≥ 0 y ≥ 0 −x − y = z 2x + 2y ≤ 8 x − y ≤ 1 x ≥ 0 y ≥ 0

(68) Solve the LP using the graphical method. minimize subject to

−x − y = z 6x + 2y ≤ 24 x − y ≤ 1 x ≥ 0 y ≥ 0

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Quadratic Business Models. (69) Write 2x2 + 3 = x2 − 2x + 4 in standard form. (Text: Section 2.1) (70) Write (y + 1)(y + 2) = 4 in standard form. (Text: Section 2.1) (71) Solve 9 − 4x2 = 0 without using the quadratic formula. (Text: Section 2.1) (73) Solve x + (74) Solve x x−1 8 x

(72) Solve x2 + 5x = 21 + x without using the quadratic formula. (Text: Section 2.1) = 9. (Text: Section 2.1)
1 . x−1

= 2x +

(Text: Section 2.1)

(75) If the proﬁt from the sale of x units of a product is P = 90x − 200 − x2 , what level of production yields a proﬁt of \$1200? (Text: Section 2.1) (76) Suppose that the proﬁt from the sale of x units of a product is P = 6400x − 18x2 − 400. Can a proﬁt of more than \$61, 800 be made? (Text: Section 2.1) (77) For y = 8 + 2x − x2 ﬁnd the vertex, determine if it gives a max or min. (78) Solve the quadratic equation 3x2 + 5x − 12 = 0. (79) Graph the quadratic function y = x2 + 4x + 3.

(80) Solve the quadratic equation 3x2 + 6x − 24 = 0. (81) Solve the quadratic equation 3x2 + 6x − 9 = 0. (82) Sketch the quadratic function y = x2 − 4.

(83) Solve the quadratic equation 3x2 + 6x − 45 = 0. (84) Sketch the quadratic function y = −x2 + 1.

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Exponential Business Models. (85) Use (a) (b) (c) (d) a calculator to evaluate 100.5 5−2.7 31/3 e2

(86) What is the future value of \$1 000 invested for 25 years at 8% compounded quarterly (4 times per year)? (87) How much interest is earned if \$32 000 is invested for 12 years at 1.5% compounded monthly? (88) A statistical study shows that the fraction of television sets of a certain brand that are still in service after x years is f(x) = e−.15x. If we start with 1 000 000 new sets, how many are in service after 10 years? How many are still in service after 11 years. (89) Which of the following is a better investment? (a) 5.0% compounded annually. (b) 4.8% compounded semi-annually. (c) 4.7% compounded quarterly. (d) 4.6% compounded monthly. (e) 4.5% compounded daily (365 days per year). (f) 4.4% compounded continuously. (90) Suppose that you deposit \$50 into a bank account paying 3% compounded monthly at the beginning of each year for 25 years. What is the future value at the end of 3 years? What is the future value after the start of the 4-th year? Solution. We ﬁrst look at the value at the end of 3 years. In this case, we have deposited \$50 at the beginning of year 1, year 2 and year 3. The ﬁrst deposit will be compounded for three years, the second for two years and the third for one year. The future value is FV = 50(1 +
.03 3∗12 ) 12

+ 50(1 +

.03 2∗12 ) 12

+ 50(1 +

.03 12 ) 12

= 50(1.0025)36 + 50(1.0025)24 + 50(1.0025)12 = 54.70257 + 53.08785 + 51.5208 = 159.3112 So, the future value at the end of three years is \$159.31. For the second scenario, we have the additional \$50 deposited at the beginning at beginning of year four. The value is \$209.31. (91) What is the present value of \$10 000 twenty years in the future if interest is 1.5% compounded daily?
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(92) What is the present value of \$10 000 twenty-ﬁve years in the future if interest is 1.3% compounded continuously? (93) If tuition fees are currently \$10, 000 per year and if they grow at 2% annually, what will they be in 20 years? (94) The purchasing power of a \$60 000 pension decreases by 3% every year. Consequently, after t years the purchasing power is P P = 60 000(1 − .03)t . What is the purchasing power after 15 years? (95) Write each of the following in exponential form. (a) 4 = log2 16 (b) 4 = log3 81 (c) 1/2 = log4 2 (d) −2 = log3 (1/9) (e) x = ln y (96) Write each of the equations in logarithmic form. (a) 25 = 32 (b) 4−1 = 1/4 (c) 53 = 125 (d) 91/2 = 3 (e) y = ex (97) Calculate the following (a) log9 2158 (b) log12(0.0195) (98) Solve the following (a) 8000 = 250(1.07)x (b) 312 = 300 + 300e−0.08x (99) Use continuous interest to approximate the doubling time of 8% compounded quarterly and 7.2% compounded monthly. (100) Suppose that the demand function for q thousand units of a commodity is given by p = 30(3−q/2 ). At what price will demand be equal to 4 000 units? How many thousands of units will be in demand if the price is \$17.32? (101) If the total cost function for a product is C(x) = e0.1x + 400 where x is the number of items produced, what is the cost of producing 30 units? If the total cost is \$404. what is the number produced? (102) Suppose that the market share y (as a percent) that a company expects t months after a new product is introduced is given by y = 40 − 40e−0.05t . What is the market share after 1 month? How long will it take for the market share to reach 24%?

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Probability Applications in Business. (103) What is the probability of obtaining exactly two heads if a fair coin is tossed three times? Solution. There are 2 × 2 × 2 = 8 possible outcomes for the three tosses and 3 ways to get two heads, namely, HHT , HT H, and T HH. The answer is 3/8. (104) What is the probability of rolling 2 distinct dice and getting a sum of 7? Solution. There are 6× 6 = 36 possible outcomes. For a sum of 7 we can have (6, 1), (1, 6),(2, 5),(5, 2),(3, 4), and (4, 3). The probability is 6/36 = 1/6. (105) What is the probability of rolling 2 distinct dice and getting a sum of 5? (106) If the odds of winning a game are 3 to 2, then the probability of winning is (107) The probability of winning a bet is 0.2. What are the odds of losing? Solution. The probability of losing is 1 − 0.02 = 0.08. So, the odds of losing are 0.8 to 0.2 or 4 to 1. (108) How many ways can 5 objects be picked from 12 distinct objects, when order is important? Solution. Since the order is important, we are asking for the number of permutations of 5 objects chosen from 12, or
12P 5

=

12! 7!

(109) A bag contains 5 red balls numbered 1 to 5 and 9 white balls numbered 6 to 14. What is the probability that a ball drawn at random is white or odd numbered? Solution. The balls that are white or odd are the twelve balls numbered 1, 3, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14. The probability of picking one of them is 12/14 = 6/7. (110) In order to win the 312 lottery grand prize you have to pick 3 numbers from 1 to 12. What is the probability of winning? Solution. The order isn’t important, so the number of ways of picking three numbers from twelve numbers is 12! 12 · 11 · 10 12C 3 = = = 220. 3!9! 3·2·1 The probability of winning is 1/220.

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(111) Given that P r(A ∩ B) = P r(B)P r(A|B) what is the probability of drawing two kings in a row from a deck of cards? Solution. Let A be the event of picking the second king and B be the event of picking 4 3 the ﬁrst king. Then, P r(B) = 4/52 and P r(A|B) = 3/51. The answer is 52 51 . (112) A company estimates that 30% of population has seen its commercial. If a person has seen the commercial, there is a 20% probability that the person will buy the product. If a person has not seen the commercial, there is a 5% probability that the person will buy the product. (a) What is the probability that a person chosen at random in the country has not seen the commercial and bought the product? (b) If a randomly selected person has purchased the product, what is the probability that the person has seen the commercial?

Solution. Let A be the event of having seen the commercial and let B be the event of having purchased the product. From the question statements we have that (a) P r(A) = .3 so that P r(notA) = 1 − .3 = .7. (b) P r(B|A) = .2. (c) P r(B|notA) = .05. (a) The question asks for P (notA∩B). We use the formula P (notA∩B) = P r(B|notA)P r(notA) = .05 ∗ .7 = .035. (b) The question asks for P r(A|B). We will use the formula P r(A|B) = P r(A∩B)/P r(B). We ﬁrst calculate P r(A ∩ B) = P r(B|A)P r(A) = .2 ∗ .3 = .06 and P r(B) = P r(A ∩ B) + P r(notA ∩ B = .06 + .035 = .095. Thus P r(A|B) = P r(A ∩ B)/P r(B) = .06/.095 = .63159. (113) Company one (C1) and company 2 (C2) are the only providers of a service in a closed market. Currently, C1 has 55% of the market. Experience from last year showed that 40% of C1 customers stayed with C1 and 50% of the C2 customers stayed with C2. (a) Given that current trends will continue, what is the transition matrix? (b) Use the transition matrix to determine the market share after two years? Solution. The transition matrix is T = The market share after one year is m1 = T m0 = The market share after two years is m2 = T m1 = .4 .5 .6 .5 .445 .4555 = . .555 .5445 .4 .5 .6 .5 .55 .445 = . .45 .555 .4 .5 . .6 .5

(114) What is the mean of {4, 5, 6, 6, 6, 7, 8, 9, 9, 10}? Solution. We add the numbers and divide by 10 to get 70/10 = 7.
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(115) What is the mode {4, 5, 6, 6, 6, 7, 8, 9, 9, 10}? Solution. The mode is the number(s) that appear most frequently. For this example, it is 6. (116) What is the median {4, 5, 6, 6, 6, 7, 8, 9, 9, 10}? Solution. The median is the number in the middle. Since there are ten numbers, we average the ﬁfth and sixth number to get the median. It is (6 + 7)/2 = 6.5. (117) What is the probability of rolling exactly four “threes ” after six rolls of a fair die? Solution. This is a binomial trial (experiment) with success being the “rolling of a three”. The probability of success is p = 1/6 and the probability of failure is q = 5/6. The number of ways of arranging four “threes ” out of six rolls is 6C 4. Since we want four successes, the 1 4 5 2 probability is 6C 4 6 . 6 (118) A company manufactures an automotive part and based on past quality surveys, it is known that the probability that a randomly chosen part is defective is 0.03. The parts are packed in boxes of twenty-four. (a) What is the probability that a randomly chosen part is good? (b) Give an expression for the probability that all parts in a randomly chosen box are good. (c) Give an expression for the probability that two defective parts are found in a randomly chosen box. Solution. (b) The probability that all parts in a randomly chosen box are good is q 24 = 0.9724 since there are twenty-four items in the box, the probability of a good part is 0.97 and the quality of the parts in the box are independent. (c) This is a binomial trial (experiment) with success being that the chosen part is defective. The probability of success is p = 0.03. Since the box has twenty-four parts, there are 24C 2 ways of having the two defective parts selected from the box. The probability is 24! (0.03)2 (0.97)22 . 2!22! (119) Suppose that a white chip is worth \$1, a green worth \$5, a blue worth \$10 and a red worth \$25. In a bag, there are 40 chips: ﬁve are white, ten are green, ﬁfteen are blue and ten are red. The experiment is to pick a single chip and the discrete random variable is x, the value of the chip. (a) In a table, give the probability distribution. (b) Calculate the expected value of the distribution.
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(a) The probability that a randomly chosen part is good is q = 1 − .03 = .97.

(c) Calculate the variance of the distribution. (d) Calculate the standard deviation of the distribution. Solution. (a) Colour Value Number Probability white 1 5 .125 green 5 10 .250 red 10 15 .375 blue 25 10 .250

(b) The expected value of the distribution is µ = E(x) = .125 · 1 + .250 · 5 + .375 · 10 + .250 · 25 = 11.375. (c) The variance is σ = (1 − 11.375)2 (.125) + (5 − 11.375)2 (.250) + (10 − 11.375)2 (.375) + (25 − 11.375)2 (.250) ≈ 70.73 (d) The standard deviation is the square root of the variance so it is σ ≈ 8.4. (e) A bag contains seven blue chips numbered 1, 2, 3, 4, 5, 6, 7 and four white chips numbered 8, 9, 10, 11. What is the probability that a chip drawn at random is blue or even numbered? (120) (Text: Section 7.4 # 21). Suppose that a box contains 3 defective transistors and 12 good transistors. If 2 transistors are drawn from the box without replacement what is the probability that (a) the ﬁrst is good and the second is defective? (b) the ﬁrst is defective and the second is good? (c) one is good and one is defective? (121) A company manufactures computer chips and, based on past experience, it is known that the probability that a randomly chosen part is defective is 0.02. The parts are packed in boxes of 10. (a) What is the probability that a randomly chosen computer chip is good? (b) Give an expression for the probability that all parts in a randomly chosen box are good. (c) Give an expression for the probability that exactly three defective parts are found in a randomly chosen box. (122) (text: Section 8.1 #18) A hospital buys thermometers in lots of 1 000. On average, one out of every 1 000 is defective. If 10 are selected at random from one lot what is the probability that: (a) all 10 are good?
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(b) exactly 5 of are good? (c) exactly 10 transistors are defective? (123) (text: Section 8.3 #48) Suppose that it has been determined that the probability that a rat injected with cancerous cells will live is 0.06. If 35 rats are injected, how many would be expected to die? The next questions come from the text book for the course on Business Data Analysis, i.e., ”Applied Statistics in Business and Economics”, Canadian Edition, by Doane, Seward, Aneja and Miller (McGraw-Hill Ryerson). (124) (Business text, Page 285, #7.18) From McDonalds.com we learn that the number of calories in an Egg McMuﬃn is a normally distributed random variable with mean 290 calories and a standard deviation of 14 calories. (a) What is the probability that an Egg McMuﬃn has fewer than 300 calories? Solution: NORMDIST (300, 290, 14, 1) = 0.76 (b) What is the probability that an Egg McMuﬃn has more than 250 calories? (c) What is the probability that an Egg McMuﬃn has between 275 and 310 calories? Solution: NORMDIST (310, 290, 14, 1) − NORMDIST (275, 290, 14, 1) = 0.78 (125) (Business text, Page 289, #7.35) The probability that a vending machine at the University of Windsor Student Centre will dispense the desired item when correct change is given is 0.90. If 200 students try the machine, ﬁnd the probabilities that (a) at least 175 will receive the desired item? and (b) that fewer than 190 will receive the correct item. Use the normal approximation to the binomial to answer the question and show the spreadsheet command (NORMDIST (N, µ, σ, 1) used to do the calculation. Solution. We have a binomial experiment with n = 200 and p = 0.9. We deﬁne a success as getting the desired item. The mean is µ = .9 ∗ 200 = 180 and the standard deviation is √ σ = 200 ∗ .9 ∗ .1 = 4.24. We approximate the binomial with the normal. The probability that at least 175 will get the correct item is The probability that fewer than 190 will get the correct item is P (X ≥ 174.5) = 1 − NORMDIST (174.5, 180, 4.24, 1) = .9. P (X ≤ 190.5) = NORMDIST (190.5, 180, 4.24, 1) = .99. Solution: 1 − NORMDIST (250, 290, 14, 1) = 0.997

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