Math Exam 2

Perform the row operation R1+R2→R1 on the matrix below. [ 6, 4 \ 3 ] [ -7, -2 \ 9 ]

6 + (−7)= -1 4 + (-2)= 2 3 + 9= 12 [ -1, 2 \ 12 ] [ -7, -2 \ 9 ]

The principal P is borrowed and the​ loan's future value A at time t is given. Determine the​ loan's simple interest rate r to the nearest tenth of a percent. P​ = ​$5100​, A​ = ​$5227.50​, t​ = 3 months

A = P (1+rt) t= 3/12 = 0.25 5227.50 = 5100 (1 + 0.25r) r = 127.50/1275 = 0.100 10%

If $2000 is invested at 3​% compounded​ continuously, what is the amount after 2 years?

A=Pert​ A= 2000e^0.03(2) A= 2123.67

What is the purchase price of a​ 50-day T-bill with a maturity value of ​$2,600 that earns an annual interest rate of 6.75​%? ​(Assume a​ 360-day year.)

t= 50 days /360 = 5/36 yr A=​P(1+​rt) 2600 = p [ 1 + 0.0675 (5/36) ] P= 2575.85

Given the annual interest rate and the compounding​ period, find​ i, the interest rate per compounding period. 17.3​% compounded daily

M= 365 I = 17.3% / 365 ≈0.047​%

Using the formula for simple interest and the given​ values, find I. P=$300​ r=7%​ t=8 years​ i=​?

i= Prt r= 7% = 7/100 = 0.07 i=Prt =​(300​)(0.07​)(8​) =​168

If​ $9000 is invested at​ 4% compounded​ quarterly, what is the amount after 6​ years?

m = quarterly = 4 i =.04/4 = 0.01 n= 4x6 = 24 A= P (1 + i)^n A= 9,000 ( 1 + 0.01 )^24 = 11427.61

Solve using augmented matrix methods 6x1 + 2x2 = 2 -9x1 - 3x2 = -3

mathway: x1 = 1/3 - x/3 x2 = 1-3x x1= 1/3 - 1/3t x2= t

Write the system of linear equations that is represented by the given augmented matrix. Assume that the variables are x1 and x2. [ 8, 9 \ 13 ] [ 1, 6 \ 5 ]

the system of linear equations is 8x1+9x2=13 x1+6x2=5.

Perform the row operation −6R2→R2 on the following matrix. [ 3, -2 \ 1 ] [ 5, -4 \ -8 ]

−6•5=−30 −6•−4=24 −6•−8=48 [ 3, -2 \ 1 ] = [ 3, -2 \ 1 ] [ 5, -4 \ -8 ] = [ -30, 24 \ 48 ]

Convert the given time period to​ years, assuming a​ 360-day year. 11 months

1. 11 months = 11 months / 12 months in a year 11 months = 11/12

A chemical manufacturer wants to lease a fleet of 28 railroad tank cars with a combined carrying capacity of 420,000 gallons. Tank cars with three different carrying capacities are​ available: 6,000 gallons, 12,000 gallons, and 24,000 gallons. How many of each type of tank car should be​ leased?

1. Argumented matrix [ 1 , 1 , 1 \ 28 ] [ 6k , 12k , 24k \ 420k ] 2. Mathway: "find reduced row echelon form" [ 1 , 0 , -2 \ -14 ] [ 0, 1 , 3 \ 42] x1 - ____ 2x3 = -14 ___ x2 + 3x3 = 42 x1 = 2t - 14 x2 = -3t + 42 x3 = t x1 ≥ 0 2t − 14 ≥ 0 t ≥ 7 x2 ≥ 0 -3t + 42 ≥ 0 t ≥ 14 7≤ t ≤ 14

Perform the row operation R1↔ R2on the matrix below. [ 1 , -9 \ 7 ] [ 8, -3 \ -2 ]

1. Flip the numbers. [ 8, -3 \ -2 ] [ 1, -9 \ 7 ]

The matrix below is the final matrix form for a system of two linear equations in the variables x1 and x2. Write the solution of the system. [ 1, 0 \ -12 ] [ 0, 1 \ 13 ]

1. Let x1 be the first variable and x2 be the second variable in the system. The equations for the final matrix are shown below. x1 + 0x2 = m 0x1 + x2 = n 2. x1 =−12, x2 = 13

The matrix below is the final matrix form for a system of two linear equations in the variables x1 and x2. Write the solution of the system. [1, -13 \ 2] [0, 0 \ 0]

1. Let x2=t. Find the value of x1. x1=13t+2 2. ​Thus, there are infinitely many solutions to the given matrix in the form x1=13t+2 ​, x2=t.

Solve the system by the substitution method. x−2y=−8 y=−5x+15

1. Substitute −5x+15 for y in the first equation. x-2y=-8 x-2(-5x+15)=-8 2. solve the resulting equation containing one variable. x−2(−5x+15)=−8 x=2 3. To find the​ y-coordinate, substitute the​ x-value into the equation y=−5x+15. Substitute 2 for x and y=−5x+15. y=5 4. the solution to the system is (2, 5) OR Use Demsos, and find the solution

A loan of $26,000 was repaid at the end of 20 months. What size repayment check​ (principal and​interest) was​ written if a 4.3% annual rate of interest was​ charged?

1. To find the amount A​ (future value) due in 20 months, use the formula A=P(1+rt) with the given values. P= 26,000 r= 4.3% = 0.043 t= 20/12 = 5/3 yrs A=P(1+rt) = 26000[ 1 + (0.043) (5/3) ] A= 27863.33

Match the system of equations below with its​ graph, and use the graph to solve the system. 5x-y = 5 3x+y = 3

1. use Desmos to find the solution. (1,0)

Solve using augmented matrix methods. 25x1 + 30x2= 5 10x1 + 12x2= 2

1. use Mathway, enter them individually or take a picture and use "solve the system of equations" 25x1 + 30x2= 5 x1= 1/5 - 6t/5 10x1 + 12x2= 2 x2= 1/6 - 5t/6 2. Thus, there are infinitely many solutions to the given system in the form x1=1/5 − 6/5t x2=t

The matrix below is the final matrix form for a system of two linear equations in the variables x1 and x2. Write the solution of the system. [ 1, -4 \ 9 ] [ 0, 0 \ 1 ]

1. use math-way, "find the determinate" there is no solution.

Solve using augmented matrices. 4x1 + 2x2 = 24 x1 - x2 = -9

1. use mathway, take a pic of both equations x1= 1 x2= 10

Convert the given time period to​ years, assuming a​ 52-week year. 11 weeks

11 weeks = 11/52 yr

Write the coefficient matrix and the augmented matrix of the given system of linear equations. 6x1 + 2x2=4 2x1 − 9x2=8 What is the coefficient​ matrix? What is the augmented​ matrix?

6x1 + 2x2=4 2x1 − 9x2=8 a. What is the coefficient​ matrix? [ 6, 2 ] [ 2, -9 ] b. What is the augmented​ matrix? [ 6, 2. \ 4] [ 2, -9 \ 8]

If ​$900 is invested at 13​% compounded ​(A) annually, ​ (B) quarterly, ​ (C) monthly, what is the amount after 10 ​years? How much interest is​ earned?

A = p (1 + i)^n a. n= # of payment periods per year x # of ​years = 10 x 1 = 10 i= r/m = 13/100 = 0.13 A= 900 (1 + 0.13)^10 = 3055.11 interest earned= 3055.11 - 900.00 = 2155.11 b. n = 4•10 = 40 i = 0.13/4 = 0.325 A= 900 (1 + 0.0325)^40 = 3234.78 interest earned= $3234.78 −​ $900.00 = 2334.78 c. n= 12 • 10 =120 i= 0.13/12 x 100/100 = 13/1200 A= 900(1 + 13/1200)^120 A= 3279.36 interest earned= 3279.36 - 900.00 = 2379.36

Suppose that the supply and demand equations for printed​ T-shirts for a particular week are given as​ follows, where p is the price in dollars and q is the quantity in hundreds. Price - supply equation: p = 0.6q + 5 Price - demand equation: p = -2.2q + 17

A)​ Find the supply and demand​ (to the nearest​ unit) if​ T-shirts are $7 each. p = 0.6q + 5 7 = 0.6q + 5 q = 3.33 = 3.33 x 100 = 333 units supplied: 333 p = -2.2q +17 7 = -2.2q +17 q = 4.55 = 4.55 x 100 = 455 units demaded: 455 Because the demand exceeds the supply at the price level of $7​, the price will go up. B) Find the supply and demand​ (to the nearest​ unit) if​ T-shirts are $12 each. p = 0.6q +5 12 = 0.6q +5 q = 11.67 = 11.67 x 100 = 1167 units supplied: 1167 p = -2.2q + 17 12 = -2.2q + 17 q = 2.27 = 2.27 x 100 = 227 units demanded: 227 Because the supply exceeds the demand at the price level of $12​, the price will go down. (C) Find the equilibrium price and quantity. 0.6q + 5 = -2.2q +17 2.8q = 12 q = 4.29 x 100 = 429 equilibrium quantity: 429 p = 0.6q + 5 p = 0.6 (4.29) + 5 p = 7.57 equilibrium price: $7.57.

An investment company pays 10​% compounded semiannually. You want to have $26,000 in the future. a. How much should you deposit now to have that amount 5 years from ​now? b. How much should you deposit now to have that amount 10 years from now​?

Formula : A= P (1+i)^n a. Compounding semiannually = 2 interest payment periods per year. n= (# of payment periods per year) x (# of​ years) = 2 x 5 = 10 i= r/m = 0.10/2 = 0.05 A= P (1+i)^n 26,000= P (1+0.05)^10 P= 1,5691.74 b. Compounding semiannually = 2 interest payment periods per year. n= (# of payment periods per year) x (# of​ years) = 2 x 10 = 20 A= P (1+i)^n 26,000= P (1+0.05)^20 P= 9799.126 = 91799.13

Use the formula for simple​ interest, I=​Prt, to find the indicated quantity. I=​$210​ P=​$6,000​ r=14​% , 0.14 t=​?

I= Prt 210=​(6,000​)(0.14​)(t) t= 0.25

Acme Annuities recently offered an annuity that pays 3.3% compounded monthly. What equal monthly deposit should be made into this annuity in order to have $500,000 in 40 years?

N = 12•40=480. I= 3.3% / 12 = 0.275 / 100 = 0.00275 FV = 500,000 PMT = FV i/(1+i)^n -1 500,000 [ 0.00275/ ( 1+0.00275)^480 -1 ] = 502.45

A sailboat costs $17,079. You pay 30% down and amortize the rest with equal monthly payments over a 6​-year period. If you must pay 8.7% compounded​ monthly, what is your monthly​ payment? How much interest will you​ pay? a. Monthly​ payments: b. ​Interest:

PMT = PV [ i / 1 - (1+i)^-n ] PV = (17,079)(0.30) = 5123.7 - 17079 = 11955.3 PMT = i = rate as decimial / # months per year = 0.087 / 12 = 0.00725 n = (#of ​years)•​(# of months per year) = 6•12= 72 a. Monthly payment: PMT= 11955.3 [ 0.00725 / 1-(1+0.00725)^-72 PMT= 213.72 b. Interest: = (72)(213.72)-(11955.3) = 3432.54

American General offers a 17​-year annuity with a guaranteed rate of 9.56​% compounded annually. How much should you pay for one of these annuities if you want to receive payments of $800 annually over the 17 year​ period?

PV = PMT [ 1-(1 + i)^-n / i ] PMT= 800 i= annually rate/# of compounding periods per year = 0.0956/1 = 0.0956 n= 17(1) = 17 = 800 [ 1 - (1 + 0.0956)^-17 / 0.0956 ] PV= 6595.858 = 6595.86

​E-Loan, an online lending​ service, recently offered 42​-month auto loans at 6.6% compounded monthly to applicants with good credit ratings. If you have a good credit rating and can afford monthly payments of $225​, how much can you borrow from​ E-Loan? What is the total interest you will pay for this​ loan? a. You can borrow $ b. You will pay a total of $

PV = PMT [ 1-(1 + i)^-n / i ] PV= PMT= 225 i= rate as decimial / # months per year = 0.066/12 = 0.0055 n= 42 a. you can borrow: = 255[ 1 - (1+0.0055)^-42 / 0.0055 ] PV = 8,417.37 b. you will pay in interest: total interest paid = (amount of all payments) - (initial loan) = (42)(225)-(8417.37) = 1032.63

If you buy a computer directly from the manufacturer for $2,100 and agree to repay it in 24 equal installments at 1.3% interest per month on the unpaid​ balance, how much are your monthly​ payments? How much total interest will be​ paid?

a. PMT = PV [ i/1-(1+i)^-n I= 1.3/100 = 0.013 2,100 [ 0.013 / 1- (1 + 0.013)^-24 PMT= 102.42 b. (24)(102.42) - (2100) = 358.08

A small plant manufactures riding lawn mowers. The plant has fixed costs​ (leases, insurance, and so​ on) of ​$120,000 per day and variable costs​ (labor, materials, and so​ on) of ​$1,700 per unit produced. The mowers are sold for ​$2,500 each. The cost and revenue equations are shown below where x is the total number of mowers produced and sold each​ day, and the daily costs and revenue are in dollars. cost equation = y= 120,000 + 1700x revenue equation = y=2500x

a. cost = revenue 120,000+1,700x = 2,500x x= 150 b. y = 2,500 (150) y = 375,000 (150, 375000) y = 120,000 + 1,700 (0) y = 120,000 y = 25000 (0) y = 0 What does the region between the lines to the left of the​ break-even point​ represent? loss What does the region between the lines to the right of the​ break-even point​ represent? profit

Use the given annual interest rate r and the compounding period to find​ i, the interest rate per compounding period. ​5.61% compounded monthly

m=12 I= 5.61/12 = 0.468

You can afford monthly deposits of $50 into an account that pays 5.0% compounded monthly. How long will it be until you have $15,000 to buy a​ boat?

n = In (1 + i (FV/PMT)) / In (1+i) The monthly interest rate​ (expressed as a​ decimal) is the annual interest rate​ (expressed as a​ percent) divided by​ 12, divided by​ 100%, or 0.004167. In ( 1 + 0.004167 x 15000 / 50 ) = 0.810975 In (1 + 0.004167) = 0.004158 n= 0.810975 / 0.004158 = 195.040

In order to accumulate enough money for a down payment on a​ house, a couple deposits $1389 per month into an account paying 8% compounded monthly. If payments are made at the end of each​ period, how much money will be in the account in 8 years?

n=12•8=96. I=8% / (12x100) = 0.0066667 FV= 1389 [ (1 + 0.0066667 )^96-1 / 0.0066667 = 1389 x 133.8686063 = 185 943.49

Write the solution of the linear system corresponding to the reduced augmented matrix. [ 1, 0 \ 7 ] [ 0, 1 \ 6 ] [ 0, 0 \ 4 ]

no solution

A small manufacturing plant makes three types of inflatable​ boats: one-person,​ two-person, and​four-person models. Each boat requires the services of three​ departments, as listed in the table. The​ cutting, assembly, and packaging departments have available a maximum of 530​, 252​, and 174 labor-hours per​ week, respectively. Construct a mathematical model to complete parts​ (A) through​ (C) below. Use​Gauss-Jordan elimination to solve the model and then interpret the solution. Dept 1-person 2-person 4-person cutting: 0.5 hr 1.0 hr. 2.0 hr assembly: 0.4 hr. 0.6 hr. 0.8 hr packaging: 0.1 hr. 0.3 hr 0.7 hr

x1 + x2 + 3x 0.5 + 1.0 + 2.0 = 530 0.4 + 0.6 + 0.8 = 252 0.1 + 0.3 + 0.7 = 174 (A) How many boats of each type must be produced each week for the plant to operate at full​ capacity? 1-person = 530 2-person = 252 4-person = 174 Augmented Matrix [0.5 , 1.0 , 2.0 \ 530] [0.4 , 0.6 , 0.8 \ 252] [0.1 , 0.3 , 0.7 \ 174] Mathway: Row Echelon Form [1 , 0 , 0 \ 60] [0 , 1 , 0 \ 140] [0 , 0 , 1 \ 180] (B) How is the production schedule in part​ (A) affected if the packaging department is no longer​ used? Augmented Matrix [ 0.5, 1.0, 2.0, \ 530] [ 0.4, 0.6, 0.8 \ 252] Mathway: Row Echelon Form [ 1, 0, 14 \ -660 ] [ 0, 1, 4, \ 860] x1 = 4t -660 x2 = -4t + 860 C) How is the production schedule in part​ (A) affected if the​ four-person boat is no longer​ produced? Augmented Matrix [0.5 , 1.0 \ 530] [0.4 , 0.6 \ 252] [0.1 , 0.3 \ 174] Mathway: Row Echelon Form [ 1 , 2 \ 1060] [ 0 , 1 \ 860] [ 0 , 0 \ -18]

Write the solution of the linear system corresponding to the reduced augmented matrix. [ 1, 0, -10 \ 1 ] [ 0, 1, 13 \ -3]

x1 - 10x3 = 1 x1 - 10t = 1 x1 = 1 + 10t x2 + 13x3 = -3 x2 + 13t = -3 x2 = -3 -13t x1 = 1 + 10t x2 = -3 -13t x3 = t

Write the solution of the linear system corresponding to the reduced augmented matrix. [ 1, 0, -10 \ 1 ] [ 0, 1, 5 \ -12 ] [0, 0, 0 \ 0 ]

x1 - 10x3 = 1 x1 - 10t = 1 x1 = 1 +10t x2 + 5x3 = -12 x2 + 5t = -12 x2 = -12 - 5t x1 = 1 + 10t x2 = -12 - 5t x3 = t

Write the solution of the linear system corresponding to the reduced augmented matrix [ 1, 0, 0 \ -4 ] [ 0, 1, 0 \ 1 ] [ 0, 0, 1 \ 0 ]

x1 = -4 x2 = 1 x3 = 0

Write the solution of the linear system corresponding to the reduced augmented matrix. [ 1 , 0 \ 1 ] [ 0, 1 \ -13] [0, 0 \ 12]

x1 = 1 x2 = −3 0 = 12

The matrix below is the final matrix form for a system of two linear equations in the variables x1 and x2. Write the solution of the system. [ 1, -2 \ 15 ] [ 0, 0 \ 0 ]

x1 = 2t + 15 x2 = t