Premium Essay

Submitted By lisajackson67

Words 1020

Pages 5

Words 1020

Pages 5

Free Essay

...Running Head: QLT1 Task 1 Rev 1 QLT1 Task 1 Rev 1 QLT1 Task 1 Mary is going away to college. In order to stay in touch with her family and friends she needs to get a cell phone with a calling plan. At this time Mary isn’t sure what her monthly phone usage is going to be. After consulting with her college friends, she estimates her usage will be between 200 and 900 minutes per month. A local phone company offers two basic calling plans: Plan A. offers a package consisting of monthly charge of $50.00 and $0.10 for each minute. Plan B. offer is a monthly fee of $15.00 and $0.18 per minute. The phone company charges for actual time used for both plans, so if Mary used .75 Min for talk time she would pay for actual time used [(.75) x per minute rate] Plan A The cost of plan (Y) X= phone usage in minutes (this is our variable) Y=50 + 0.10(x). The cost of plan A (Y) is $50 base charge plus $.10 per each minute (x) represented in a linear equation. Plan B X= phone usage in minutes (this is our variable) Y=15 + .18(x). The cost of plan B (Y) is $15 base charge plus $.18 per each minute (x) represented in a linear equation. 1. Which plan would offer greater savings if Mary’s monthly phone usage was 200 min? Plan A Cost of phone plan after using 200 minutes. Y=50 + .10(x) X 200 Y=50 + .10(200) Y=50 + 20 Y=70 Plan A would cost Mary $70 per month after 200 minutes usage Plan B Cost of phone plan after using 200......

Words: 659 - Pages: 3

Free Essay

...daycares to determine the best option for their 1 year old son. A home-based daycare charges a flat rate of $5 per hour. A center-based daycare charges a fixed rate of $185 per week, providing 40 hours of childcare. Above 40 hours, the center-based daycare then charges a fixed rate of $8 per hour. The couple determines that the driving distance to each daycare is the same, thus driving expenses do not need to be considered. The couple will require 50 hours of childcare per week and are looking for the cheapest daycare, as they are soon expecting their second child. Part B: 1. “y” represents total cost per week in dollars “x” represents hours Home based daycare: y=5x Center based daycare: y=185+8(x-40), x≥40 hours 2. For the home based daycare, they charge a flat rate of $5 per hour. Thus the couple can calculate their cost by multiplying $5 by the number of hours they need for childcare each week. For the center based daycare, the couple can calculate their weekly cost by adding the $185 fixed rate to the cost of any additional hours needed after 40. Thus, if they are going over 40 hours, they would subtract 40 hours from the total number of hours needed and multiply that by the $8 the center charges for each additional hour. 3. y=5x y=185+8(x-40) y=185+8(x-40) 5x=185+8x-320 5x=-135+8x 5x-8x=-135+8x-8x -3x=-135 -3 -3 x=45 y=5x y=5(45) y=225 Thus, both the center based and the home based daycares will charge $225 for 45......

Words: 495 - Pages: 2

Free Essay

...QLT1 Task 5 A. Create a story problem using one of the above real-world scenarios as a basis, including realistic numeric values, by doing the following: 1. Describe the real-world problem. I was looking into phone plans and stumbled upon T-Mobile, and I decided that I needed a cell-phone and took a look at the plans. T-mobile had one plan that was 50 dollars a month and is unlimited talk, text and web, T-Mobile also has a plan for 30 dollars a month for 1,500 talk and text minutes. After you go through your allotted 1,500 talk and text time was up, the cost skyrockets up to 10 cents a minute. 10 cents a minute comes out to 6.00 per hour, I thought in my head. I decided that instead of jumping into a decision about phone plans, that I should first go home and do the math. I wanted to figure out which plan was going to be the most cost effective, and which plan would suite my needs the best. 2. Explain all needs (e.g., financial, non-financial, situational) of the hypothetical consumer. The needs include many different factors: • The first and most important factor how much do you use your phone? The breaking point on the problem today is 200 talk or text above the 1500 minute plan and I will be paying less up front but more on the back end which is no good. If I chose the 1500 minute plan, I will not want to be going over the 1500 minutes as then the cost would go up to 10 cents a minute or 6.00 per hour. • Is this replacing a work or office phone? ......

Words: 1158 - Pages: 5

Premium Essay

...TASK 5 Databases a) A database is a structured collection of records or data that is stored in a computer system. b) Databases are used by storing data so it can be quickly accessed, varied information could be kept about anything, from students to stock e.g. computers and equipment, you could use MS Access to make a Database, you can use queries to find things quickly. It would be best to create a database in MS access because things can be changed and added. For instance you could add a new field or put in some new data. It is also easy to use in MS Access because it sets it out as a table so it’s easy to edit and read. c) College could use this to keep records on all their students and teachers, they could have all personal details, like Phone numbers, address, and how well their doing in their course, it could keep track of all their absences. It could hold all information about people who are visiting the college, and when things are going to be organised like student days and open days. Internet a) The internet is a server where people can add web pages to show information; you need an internet connection to be able to access the internet. b) The internet is used for browsing web pages, I would use the internet for going on face book, but a business would use the internet for things like buying stock, or advertising there business to people, some business may also have their own website to advertise. A college may have a website to describe about all the......

Words: 757 - Pages: 4

Free Essay

...x 0 45 30 Y = (-2/3)x + 30 30 0 10 Y intercept: (0,30) Y = -2/3(x) + 30 Y = 0 + 30 Y= 0 X intercept: (45,0) 0 = -2/3 (x) + 5 0 – 30 = -2/3 (x) +30 -30 -30 = -2/3 x 2/3(x) = 30 X = 30 ● 3/2 X = 45 Height of the beam 30 ft. away from the face of building is 10 ft. Y = -2/3 (x) + 30 Y = -2/3 30 + 30 Y = -20 + 30 Y = 10 Y represents the height above the ground in feet. 35 Y intercept (0, 30) 30 25 The graph depicts a visual representation of the path of the laser beam using only quadrant I. Quadrants ll, lll, and lV are not needed to show the direction of the laser beam. Y = (-2/3)x + 30 Y = (-2/3)x + 30 20 When the laser beam is shined at the ground at 30 ft. up it will hit the ground at 45 ft. from the face of the building using the equation Y=(-2/3)x + 30. 15 10 (30, 10) 5 X Axis 0 0 Y Axis 10 20 30 The height of the beam 30 ft. away from the face of the building is 10 ft. X intercept (45, 0) 40 50 X represents the distance from face of bldg. in ft. The laser beam hits the side of the building at 30 ft. which is the Y intercept (0, 30) and the laser beam hits the ground at 45 ft. from the face of the building at the X intercept (45, 0). X intercept (45, 0). ...

Words: 274 - Pages: 2

Free Essay

...Task 1 1. Graph the points (1,0,-6,3/4,1.7) in single number line. 2. Graph the following points on a single coordinate plane. Make sure to include labels for each quadrant of the coordinate plane. • Point 1: (3, –2) • Point 2: (0, 0) • Point 3: (–1, 7) • Point 4: (3, 5) • Point 5: (–4, –5) 3. Graph the following functions on separate coordinate planes. Function 1: y= 2x − 1 If x = 0 y= 2x − 1 y= 2(0) − 1 y= − 1 If x = 1 y= 2x − 1 y= 2(1) − 1 y= 2 − 1 y= 1 If x = 2 y= 2x − 1 y= 2(2) − 1 y= 4 − 1 y= 3 If x = 3 y= 2x − 1 y= 2(3) − 1 y= 6 − 1 y= 5 If x = 4 y= 2x − 1 y= 2(4) − 1 y= 8 − 1 y= 7 Function 2: y= (–3/4)x+ 5 If x = 0 y= (–3/4)x+ 5 y= (–3/4)(0)+ 5 y= 5 If x = 1 y= (–3/4)x+ 5 y= (–3/4)(1)+ 5 y= –3/4 + 5 y= 4 1/4 If x = 2 y= (–3/4)x+ 5 y= (–3/4)(2)+ 5 y= –1 1/2 + 5 y= 3 1/2 If x = 3 y= (–3/4)x+ 5 y= (–3/4)(3)+ 5 y= –2 1/4 + 5 y= 2 3/4 If x = 4 y= (–3/4)x+ 5 y= (–3/4)(4)+ 5 y= –3 + 5 y= 2 Function 3: y= x2 – 4 If x = 0 y= x2 – 4 y= (0)2 – 4 y= -4 If x = 1 y= x2 – 4 y= (1)2 – 4 y= 1 – 4 y= -3 If x = 2 y= x2 – 4 y= (2)2 – 4 y= 4 – 4 y= 0 If x = 3 y= x2 – 4 y= (3)2 – 4 y= 9 – 4 y= 5 If x = 4 y= x2 – 4 y= (4)2 – 4 y= 16 – 4 y= 12 Function 4: y= –3x2 − 6x – 5 If x =...

Words: 458 - Pages: 2

Free Essay

...000278346/Arpe/QLT1 Task 212.1.2: Task 1 A. Complete the following graphs: 1. Graph the following values on a single number line. Value 1: 1 Value 2: 0 Value 3: -6 Value 4: ¾ Value 5: -1.7 B C E D A Page 1 of 6 000278346/Arpe/QLT1 Task 212.1.2: Task 1 2. Graph the following points on a single coordinate plane. Make sure to include labels for each quadrant of the coordinate plane. Point 1: (3,-2) Point 2: (0.0) Point 3: (-1,7) Point 4: (3,5) Point 5: (-4,-5) (-1,7) (3,5) (3,-2) ) (0,0) (3,-2) (-4,-5) Page 2 of 6 000278346/Arpe/QLT1 Task 212.1.2: Task 1 3. Graph the following functions on separate coordinate planes. In each graph, label each axis of the coordinate plane. Additionally, label each intercept as “x-intercept” or “y-intercept” and include the ordered pair. Whenever applicable, label the vertex as “vertex” and include the ordered pair Function 1: y = 2x – 1 (3,5) (2,3) ( ½ ,0) (0,-1) x-intercept y-intercept (-1,-3) Page 3 of 6 000278346/Arpe/QLT1 Task 212.1.2: Task 1 Function 2: y = (-3/4)x + 5 (-4,8) x-intercept (0,5) (4,2) y-intercept (6 ½,0) (8,-1)) Page 4 of 6 000278346/Arpe/QLT1 Task 212.1.2: Task 1 Function 3: y = x2 – 4 (-3,5) (3,5) (-2,0) x-intercept (2,0) x-intercept (-1,-3) (1,-3) (0,-4) Vertex Page 5 of 6 000278346/Arpe/QLT1 Task 212.1.2: Task 1 Function 4: y = -3x2 – 6x –......

Words: 295 - Pages: 2

Free Essay

...A1) A parent is looking for the best option for daycare for a child. A home-based option charges a flat rate of $10 per hour. A center-based option charges a fixed fee of $332.50 per week for 35 hours and then a fixed rate of $11 per hour for any hour provided over the set amount. Determine which option is most advantageous for the parent based upon the parents’ needs. A2) The parent works a 40 hour work week. Each of the daycare options is a half an hour away from the parent’s office. This means that the child is there for a total of 45 hours a week. The parent has federal holidays (10 holidays in 2014), the day after Thanksgiving, and Christmas Eve off so there will be weeks that the child will not be in daycare for the full week. The parent also receives two weeks of vacation each year. Other situations to consider are that the child and/or parent could get sick or they could take a short vacation and the child will not have to attend the daycare for the full week. A3) If the parent chooses to go with the home-based option, the parent will be charged a flat rate of $10 per hour. This option gives the parent the freedom to either take the child to daycare or to not and only be charged for the hours the child is there. If the child is there for 45 hours a week, the parent will pay $450 for that week; if the child is there for only 20 hours a week, the parent will pay $200 for that week. If the parent chooses to go with the center-based option, the parent will be charged......

Words: 837 - Pages: 4

Premium Essay

...QLT1 Task 5 Daycare Cost Analysis A.) Basic information: The Depp family is seeking child care for their 2 year old son Jack. The family needs care for approximately 42-50 hours per week. They have interviewed several care providers and have selected two providers in the vicinity of their employers and now need to decide between: one is family day care home that charges $5 per hour per child and the other is a child care center that charges $200 per week for 40 hours with an additional $10 per hour over 40 hours. Additional factors considered are that both facilities provide lunch and snacks, other facilities did not; as well the center will provide a 25% discount on a second child. Although the Depps have only one child however will be adding to their young family. 1.) Rates: i. Daycare rates: Home care $5/hour ii. Center $200/week for 40 hours + $10/hour over 40 hours 2.) Other expenses iii. Meals included with both facilities iv. Both located within 10 blocks of employer, transportation costs are negligible 3.) Other considerations v. Parents are state employees 37.5 hours per week + 1 hour lunch = 42.5 hours @ employer per week minimum vi. Currently have one child in need of care, but is considering another child in the next 1-2 years vii. Family day care home is 1 employee, no coverage for sick days/vacation; center has multiple staff members to......

Words: 456 - Pages: 2

Free Essay

...QLT1 – Task 1 Competency 212.1.2: Solving Algebraic Equations A. 1. Graph the following values on a single number line • Value1: 1 • Value 2: 0 • Value 3: -6 • Value 4: ¾ • Value 5: -1.7 [pic] 2. Graph the following points on a single coordinate plane. Make sure to include labels for each quadrant of the coordinate plane. • Point 1: (3,-2) • Point 2: (0,0) • Point 3: (-1,7) • Point 4: (3,5) • Point 5: (-4,-5) Quadrant II Quadrant I [pic] Quadrant III Quadrant IV 3. Graph the following functions on separate coordinate planes. • Function 1: y= 2x − 1 y=2(0) – 1 0=2x – 1 y= 0 – 1 +1 +1 y= -1 1 = 2x /2 /2 ½ = x [pic] • Function 2: y= (–3/4)x + 5 y=(-3/4)0 + 5 0=(-3/4)x + 5 y= 0 + 5 -5 -5 y = 5 (0,5) -5 = (-3/4)x *4 *4 -20 = -3x /-3 /-3 6.67 = x (6.67,0) [pic] • Function 3: y=x2 – 4 y=02 – 4 0=x2 – 4 y=(-2) 2 – 4 y=(3)2 -4 y=(-3)2-4 y=-4 +4 +4 y = 4 – 4 y=9-4 y=9-4 4= x2 y = 0 y = 5 y = 5 √4 = √x2 2 = x [pic] • Function 4: y= –3x2 – 6x – 5 If x=0 then If x=-1 then y= –3(0)2 – 6(0) – 5 y= –3(-1)2 – 6(-1) – 5 y = 0 – 0 – 5......

Words: 424 - Pages: 2

Free Essay

...A. 1. A. 2. Savings Plan A y= $20x + $400 10x+600=20x+400 -10x-400=-10x -400 200=10x /10 /10 20=x (# 0f months) Savings PlanB y= $10x + $600 y=10(20)+600 y=200+600 y=800 ($ saved) Savings Plan A X-intercept y=0 0=20x+400 -400=20x -400/20=x -20=X Savings PlanB X-intercept y=0 0=10x+600 -600=10x -600/10=x -60=x Solution A. 3. Both plans yield identical balances after 20 months at $800. Months Plan B Plan A -60 0 -20 0 0 600 0 400 2 620 2 440 Solution 3 630 3 460 1000 4 640 4 480 (20,800) y-intercept Quadrant II 5 650 5 500 800 (0,600) 6 660 6 520 600 Quadrant I 7 670 7 540 400 8 680 8 560 x-intercept (0,400) y-intercept x-intercept 200 9 690 9 580 (-60,0) (-20,0) 10 700 10 600 0 -70 -60 -50 -40 -30 -20 -10 0 10 20 30 11 710 11 620 X-Axis Months 12 720 12 640 13 730 13 660 Plan B y=10x+600 Plan A y=20x+400 14 740 14 680 Linear (Plan B y=10x+600) Linear (Plan A y=20x+400) 15 750 15 700 16 760 16 720 17 770 17 740 A.3a1. Plan B yields more at $740 vs. Plan A at $680. 18 780 18 760 A.3ac. Plan A yields more at $860 vs Plan B at $830. 19 790 19 780 A.4. The solution lies in quadrant 1 where the savings 20 800 20 800 begins until the intersection at (20,800). 21 810 21 820 x & y coordinates are all positive from month 1 through month 20 22 820 22 840 which means its Quadrant 1. 23 830 23 860 Y-Axis (Total Savings in $) ...

Words: 320 - Pages: 2

Premium Essay

...A1 Andrew has just moved to Los Angeles and bought a car and is looking for the best parking option that will allow him to save the most money. There are two parking garages by his work. One charges $40 per month and the other charges 20 cents per hour. Which option will be less costly for him? A2 Since Andrew is new to the area it is necessary that he find the cheapest parking option possible in order to cut down on expenses in the new city as he has just purchased a new car and had high moving costs. A3 Andrew’s two options are the monthly parking garage which charges $40 per month or the hourly parking garage which charges 20 cents per hour. B1 A work day has 8 hours A is the number of days the car will be parked. T will be the total price Monthly parking garage: (A-A)+40=T Hourly parking garage: (0.2 x 8)A=T B2 The monthly parking garage doesn’t charge by the day so having the days at zero just leaves the $40 charge. The hourly parking garage multiplies the rate by 8 hours since there are 8 hours in a work day. That is then multiplied by the total number of days in a month that the car will be parked. B3 We set both equations equal to each other in order to find the solution for the number of days. Set both equations equal to each other (A-A)+40=(0.2 x 8)A Subtract (A-A) 40=(0.2 x 8)A Multiply inside the parentheses (0.2 x 8) 40=(1.6)A Multiply (1.6)A 40=1.6A Divide both sides by 1.6 25=A A=25 days solution We...

Words: 501 - Pages: 3

Premium Essay

...Task 5 A. Jenny is looking for the best option for daycare for her son. She lives in a small town, so her options are limited to two daycare centers. Option A is a home-based facility which charges $9.00 per hour that the child is at the home. Option B is a center-based facility which charges $150 for the first 20 hours and then charges $10 per hour thereafter. Due to a new promotion, Jenny will be required to work more hours away from home. Which facility will cost less for Jenny to have her son in daycare for a work week (40 hours)? B. I. h = how many hours of child care is needed = 40 f = the flat hourly rate charge t = total amount due for the week Option A: f * h = t a. 9 * 40 = $360 Option B: 150 + f(h-20) = t, for h >=20 b. 150 + 10(40-20) = $350 150 = t, for h < 20 c. 150 = total II. I set the hours to "h", the hourly fee to "f" and the total to "t". a. Option A: This option only has an hourly fee for each hour. The fee is $9 and the amount of hours is 40 so I multiplied the hours by the fee. b. Option B: This option has a flat rate of $150 for the first 20 hours and then an hourly rate of $10 thereafter. For the first equation that refers to over 20 hours, I subtracted 20 hours from the total time, multiplied it by $10, and then added it to $150. For the second equation that refers to any amount less than 20 hours, the total is always......

Words: 531 - Pages: 3

Premium Essay

...Task 5 Moving Truck Cost Analysis The Thompsons have lived in Canada all their life but recently Mr. Thompson got a job offer he could not refuse. The new job will require they move to the west coast of the United States. This is going to require they move across the country from New Brunswick, Canada to Seattle, WA. Which is a total of 3200 miles. Although the company Mr. Thompson will be working for will reimburse them for the move, they still have a budget they need to stay under so they can have enough money for hotel stays, stocking the house with food, cleaning supplies etc. when they arrive. They are planning on leaving early and site seeing on the way but still arrive at least two weeks early so they have time to settle in. Arriving early though means there will be a big gap between paychecks so budgeting is very important. They narrowed it down to two companies, Acme Moving which charges $2.00 per mile and Lightening Moving that charges $1500 for the first thousand miles and $2.25 for each mile over 1000. They have to analyze the costs and figure out which one is going to be the best deal. The Equation: Lightening Moving Co. Lightening Moving charges $1500 for the first 1000 miles and then $2.25 for each mile over 1000. Let X = Amount of miles traveled 1500 = the initial cost of the first 1000 miles 3,200 = the total amount of miles traveled $2.25 per mile over 1000 miles therefore; 1500 + 2.25(X-1000) = Cost of moving over 1000 miles 1500 + 2...

Words: 550 - Pages: 3

Premium Essay

...Analysis for Business (QAT1) Submitted 05/05/2015 Assignment 309.3.3-04 Version LMF5-28 Student: Richard McClanahan Student ID: 000343792 TASK #5 Answer Task 5A Calculate the expected value for EACH of the four decision branches. 1. Develop Thoroughly: GOOD) $500,000 (0.45) = $225,000 MOD.) $25,000 (0.10) = $2,500 POOR) $1,000 (0.45) = $450 TOTAL EXPECTED VALUE: $227,950 2. Develop Rapidly: GOOD) $500,000 (0.52) = $260,000 MOD) $25,000 (0.23) =$5,700 POOR) $1,000 (0.25) =$250 TOTAL EXPECTED VALUE: $265,950 3. Strengthen Products GOOD) $2,000 (0.33) = $660 MOD) $10,000 (0.52) = $5,200 POOR) $3,000 (0.15) = $450 TOTAL EXPECTED VALUE: $6,310 4. Reap without investing GOOD) $10,000 (0.33) = $3,300 POOR) $1,000 (0.67) = $670 TOTAL EXPECTED VALUE: $3,970 EXPLINATION: We take the projected payoff and multiply that payoff by the probability factor. So if the good payoff to develop a product rapidly is $500,000, we then multiply that by the probability factor of 52%, or 0.52. That gives us a probable payoff of $260,000. Following this simply process, we extrapolate these results as listed above. ANSWER TASK 5B After calculating the total expected value for each decision alternative, the most profitable decision would be to RAPIDLY DEVELOP new products for a probable payoff......

Words: 364 - Pages: 2