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# employee has worked 1062.45 hours and receives 0:02 hours per hours worked for sick leave

employee has worked 1062.45 hours and receives 0:02 hours per every hour worked  for sick lease.  What is his sick leave hours?

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Start studying 2010 CPP Review ... a company's production workers receive \$0.32 for each unit ... at a regular rate of \$11 per hour. The employee worked 2 ...

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\$9.10 per hour. He worked 38 hours last week. ... gross pay does Albert receive for 4 weeks of work? ... Employee Hourly Pay Hours Worked Gross Pay

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Calculate the total number of hours of sick leave the employee has ... also receive 2 weeks of sick leave per year but ... work out the average hours worked per

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The rules for the P46 form have changed and employers' HR and ... not the number of hours worked. ... allows and employee to receive £6479 per annum ...

### Convert Days to Hours - CalculateMe.com

Convert Days to Hours,Time Conversions. ... 0.01: 0.24: 0.02: 0.48: 0.03: 0.72: 0.04: 0.96: 0.05: 1.2: 0.06: 1.44: 0.07: ... 45: 1,080: 46: 1,104: 47: 1,128: 48 ...

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Evaluation of an 8 hour versus a 12 hour shift roster on employees at ... The amount of annual leave and sick leave taken by the employees as ... 0: Hours worked per ...

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I work a temporary job for £10.50 per hour for 30 hours a week with no sick pay ... they worked 41 hours per ... 5 hours to 35hours.I have received pay rise ...

### Retirement - Wikipedia

... A person may also semi-retire by reducing work hours. ... 45%: 12%: 1%: 0%: Cambodia: 50: 55: 16%: 1%: 0%: 0%: Denmark: ... Using i real =0.02, or 2% per year ...

### Calculating Time in Excel • My Online Training Hub

Calculating time in Excel can be ... hours worked for my employees. 42:30:00 and 21:02:12 and 12 ... and have worked out the hours per ...

## Suggested Questions And Answer :

### employee has worked 1062.45 hours and receives 0:02 hours per hours worked for sick leave

start: werk 1062.45 ours & get sik pae=0.02 \$ per our sik pae=0.02*1062.45=21.249 round2(21.249)=21.25\$

### how long will it take each pipe alone to fill the tank?

Since it takes 100 hours to fill the tank when both pipes are used together, in 1 hour, 1/100 of the tank will be filled. Pipe A contributes 1/x of the tank in one hour and pipe B contributes 1/y of the tank. So the two fractions added together fill 1/100 of the tank in an hour: 1/x+1/y=1/100. Since pipe A fills 1/x tank in an hour, it would take it x hours to fill the tank by itself. Similarly, pipe B would take y hours. Pipe A fills 6/x of the tank in 6 hours, leaving 1-6/x of the tank to be filled. Since pipe B takes y hours to fill the tank, it would take fewer hours if the tank was already partly filled. So, by proportion, it would take y(1-6/x) hours to top up the tank by itself. We know this time is 18 hours, so y(1-6/x)=18. We have two equations in x and y, so we can work out the two variables. But there's something wrong! If A and B working together take 100 hours, it surely takes longer to fill the tank working in relay. So I think that 100 hours should be 10 hours. If we proceed as the question indicates, we'll see that we have a pipe removing rather than adding water to the tank.  Multiply the first equation through by 100xy: 100y+100x=xy. Multiply the second equation through by x: xy-6y=18x. Therefore xy=18x+6y=100y+100x, so 94y+82x=0 and x or y would be negative. Let's replace 100 with 10: 18x+6y=10x+10y, 8x=4y so y=2x. 10x+20x=2x^2 (substituting for y in 10x+10y=xy). So x^2=15x, and x=15 (we disregard x=0). Therefore y=30. Check: 1/15+1/30=3/30=1/10 and it takes pipe A 6 hours to fill 6/15=2/5 of a tank, leaving 3/5 left to fill. This will take pipe B 3/5 of 30 hours=18 hours to complete. So with the adjustment, x=15 hours and y=30 hours.

### Shawn took in \$69.15 in two hours work on Saturday selling Belts for \$8.05 and earrings for \$4.50. How many of each did Shawn sell?

Shawn took in \$69.15 in two hours work on Saturday selling Belts for \$8.05 and earrings for \$4.50. How many of each did Shawn sell? I'm not too sure how ​you are expected to solve this problem. You will end up with a Diophantine Equation (one involving integer coefficients and integer solutions only) If you have never heard of Diophantus, then the precise mathematical solution later on should be ignored. Instead ...   The (Diophatine) equation is 161B + 90E = 1363 You should be able to show that Shawn can only sell a maximum of 8 belts (i.e. if he sells no earrings), and a maximum of 15 earrings (i.e. if he sells no belts). So start with the smaller number (8). Shawn sells up to 8 belts. So put B = 1 in the (Diophantine ) equation and see if B is an integer. If not, then B=1 cannot be a solution, So now try again with B = 2, and so on ..     My precise mathematical solution now follows. Let B be the number of belts sold @ \$8.05 per belt. Let E be the number of earrings sold @ \$4.50 per pair. Total money made is P = \$69.15 Total money made is: B*\$8.05 + E*\$4.50 Equating the two expressions, 8.05B + 4.50E = 69.15 Making the coefficients as integers we end up with a Diophantine equation. 161B + 90E = 1383 We now manipulate the coefficients 161 and 90. We should end up with a relation between the two that should produce the value 1. This manipulation is in two parts. In the first line, we are writing the larger coefft, 161, as a multiple of the smaller coefft plus a remainder. 1st Part 161 = 1x90 + 71  90 = 1x71 + 19     now we write the 1st multiple as a multiple of the 1st remainder + remainder 71 = 3x19 + 14    we write the 2nd multiple as a multiple of the 2nd remainder + remainder 19 = 1x14 + 5 14 = 2x5 + 4 5 = 1x4 + 1 ============ 2nd Part We rewrite that last equation so that 1 is on the left hand side 1 = 5 – 1x4 We now use the remainder from the next equation up. 1 = 5 – 1x(14 – 2x5) 1 = 3x5 – 1x14 And again we use the remainder in the next equation up 1 = 3(19 – 1x14) – 1x14 1 = 3x19 – 4x14 1 = 3x19 – 4(71 – 3x19) 1 = 15x19 – 4x71 1 = 15(90 – 1x71) – 4x71 1 = 15x90 – 19x71 1 = 15x90 – 19(161 – 1x90) 1 = 34x90 – 19x161 ================ Therefore, 1383 = (-19x1383).161 + (34x1383).90 1383 = -26,277*161 + 57,022*90 Which implies B = -26,277 E = 47,022 which is obviously incorrect! But all is not lost. These two values for B and E are simply two values that happen to satisfy the original Diophantine equation. What we need is the general solution, which now follows. In general, B = -26,277 + 90k    (using the coefft of E) E = 47,022 – 161k    (using the coefft of B) If you substitute these expressions into the original Diophantine equation, the k-terms will cancel out, leaving you with the original eqn. We know that B and E must be smallish numbers and that neither can be negative. For example, if no earrings were sold, Shawn would need to sell over 8 belts to clear \$69, and if no belts were sold, then Shawn would need to sell over 15 earrings. So Shawn needs to sell between 0 and 8 belts and between 0 and 15 earrings. Setting k = 292, B = -26,277 + 90*292 = 3 E = 47,022 – 161*292 = 10 B = 3, E = 10 If k is greater than or less than 292 by 1, or more, then either B will be negative or E will be negative. So this is the answer. Check 3*8.05 + 10*4.50 = 25.15 + 45.00 = 69.15 – Correct! Answer: Shawn sells 3 belts and 10 earrings

### Time Problem

Let's say the painter takes p hours to paint the house on his own, then his brother takes p+3 hours. Let's suppose the house needs P litres of paint to paint it. The paint rate for the painter is P/p litres per hour; his brother's rate is P/(p+3) l/hr. If the house takes 6 2/3=20/3 hrs to paint when they work together then the paint rate is P÷20/3=3P/20. When we add the paint rates together we have P/p+P/(p+3)=3P/20 so we can cancel P out leaving 1/p+1/(p+3)=3/20. Multiply through by 20p(p+3): 20p+60+20p=3p(p+3)=3p^2+9p; 3p^2-31p-60=0=(3p+5)(p-12). We only need the positive root p=12 hours. The painter takes 12 hours to paint the house on his own and his brother takes 15 hours.

### \$8 per hour mowing lawns and & \$12 per hour babysitting. she wants at LEAST \$100 per week but can't won't more than 12 hours a weeks write a graph & system of lines inequalities

\$8 per hour mowing lawns and & \$12 per hour babysitting. she wants at LEAST \$100 per week but can't won't more than 12 hours a weeks write a graph & system of lines inequalities Mowing a lawn = \$8/hr Babysitting= \$12/hr Work Time = WT <= 12 hrs/wk Income = M >= \$100 / week Let x be the number of hrs mowing lawns.   “   y  “    “         “         “    “   babysitting. Setting up our inequalities, M = 6x + 12 y   à 8x + 12y >= 100,  i.e. 2x + 3y >= 25 WT = x + y     --> x + y <= 12 The system of inequalities is, 2x + 3y >= 25 x + y <= 12 x >= 0, y >= 0 These inequalities can be graphed using the lines, l1: 2x + 3y = 25 l2: x + y = 12 and are shown below, The inequality x + y <= 12 is given by all that area below the line l1, such that both x and y are positive, i.e. the triangle AOB. The inequality 2x + 3y >= 25 is given by all that area above the line l2, such that both x and y are positive, i.e. the triangle COD. The common area satisfying both requirements is the triangle ACE. Only points within the triangle ACE are both below l1 and above l2. Income, M, is maximised for that point within triangle ACE that is furthest from the origin, which is the point A, i.e. where x = 0 and y = 12. M = 6x + 12y 6*9 + 12*12 = 144 Max income: \$144

### 4 poems by Shakespeare, 5 poems by Coleridge, 2 poems by Tennyson,

20. The two box-and-whisker plots below show the scores on a math exam for two classes. What do the interquartile ranges tell you about the two classes? (1 point) (0 pts) Class A has more consistent scores. (1 pt) Class B has more consistent scores. (0 pts) Overall class A performed better than class B. (0 pts) The lowest score was in class B. 1 /1 point 21. Is the following data set qualitative or quantitative? the numbers of hours spent commuting to work by the employees of a company (1 point) (0 pts) qualitative (1 pt) quantitative 1 /1 point The item below has been reviewed and is scheduled to be updated. All students will receive full credit for any response to the following. 22. Identify the sampling method. You want to determine the number of text messages students at your school send in a month. You randomly ask students in each of your classes. (1 point) (1 pt) random (0 pts) systematic (0 pts) stratified (0 pts) none of these 1 /1 point 23. A mathematics journal has accepted 14 articles for publication. However, due to budgetary restraints only 7 articles can be published this month. How many specific ways can the journal editor assemble 7 of the 14 articles for publication? (1 point) (0 pts) 14 (0 pts) 3,432 (0 pts) 98 (1 pt) 17,297,280 1 /1 point 24. Your English teacher has decided to randomly assign poems for the class to read. The syllabus includes four poems by Shakespeare, five poems by Coleridge, two poems by Tennyson, and two poems by Lord Byron. What is the probability that you will be assigned a poem by Coleridge, and then a poem by Lord Byron? (1 point) (0 pts) (0 pts) (1 pt) (0 pts) 1 /1 point The final score is 24/24 (100%).

### what is \$60 minus 2/5

to do this problem first we find out how much he earns an hour says he ears a day so assume 24 hours he works lol 60\$/24 = 2.5\$ an hour   now we find 2/5 of  day which is 2/5 x 24 = 9.6 hours x 2.5\$ = \$24 he earned \$24 for this sick day.

### The temperature in a greenhouse from 7:00p.m. to 7:00a.m. is given by f (t)= 96 - 20sin (t/4), where f (t) is measured in Fahrenheit, and t is the number of hours since 7:00 p.m.

The temperature in a greenhouse from 7:00p.m. to 7:00a.m. is given by f (t)= 96 - 20sin (t/4), where f (t) is measured in Fahrenheit, and t is the number of hours since 7:00 p.m. A) What is the temperature of the greenhouse at 1:00 a.m. to the nearest Fahrenheit? (B) Find the average temperature between 7:00 p.m. and 7:00 a.m. to the nearest tenth of a degree Fahrenheit. (C) When the temperature of the greenhouse drops below 80 degreeso Fahrenheit, a heating system will automatically be turned on a maintain the temperature at minimum of 80 degrees Fahrenheit. At what values of the to the nearest tenth is the heating system turned on? (D) The cost of heating the greenhouse is \$0.25 per hour for each degree. What is the total cost to the nearest dollar to heat the greenhouse from 7:00 p.m. and 7:00 a.m.?   The equation is: f(t) = 96 – 20sin(t/4), 0 <= t <= 12 (A)  At 1.00 a.m. t = 6 f(6) = 96 – 20.sin(6/4) = 96 – 20*0.99749 = 96 – 19.9499 f(6) = 76 ⁰F (B)  The average temperature would need to be worked out by sampling the temperature at different times throughout the night. Divide the temperature range into N equal intervals, giving N+1 sampling points. We would then have T1 = f(δt), T2 = f(2δt), T3 = f(3δt), ... ,Tn = f(nδt) Where δt = range/N = 12/N, and n = 0..N Giving Tn = 96 – 20.sin((12n/N)/4) = 96 – 20.sin(3n/N) Then Tav = (1/N)*sum(Tn, n = 0 .. N) i.e. Tav = (1/N)*sum(96 – 20.sin(3n/N), n = 0 .. N Tav = 96 – 20. (1/N)*sum(sin(3n/N), n = 0 .. N I used Maple to evaluate the above summation. The results are tabulated as follows.                                  Average Temperature Num Intervals            3       4        6       10       20      50     100    200 Tav over the range 83.39 83.01 82.78 82.69 82.69 82.71 82.72 82.73 As can be seen from the table the temperature is averaging out at:  Tav = 82.7 ⁰F (C)  T = f(t) = 96 – 20sin(t/4), 0 <= t <= 12 At T = 80 ⁰F,            96 – 20sin(t/4) = 80 20.sin(t/4) = 96 – 80 = 16 sin(t/4) = 0.8 t/4 = 0.927295 t = 3.70918 t = 3.7 (to nearest tenth) (D)  The temperature will (normally) drop to 80 ⁰F after t = 3.7 hours and rise again to 80 ⁰F when t = 12 – 3.7 = 8.3 hours. Heating system is turned on for 8.3 – 3.7 = 4.6 hours Cost of heating is 4.6*80*0.25 = 4.6*20 = 92 Cost = \$92

### It takes Kevin 3 hours longer than Walter to build a tree house.Together they can do the job in 2 hours. How long would it take each man to build the tree house on his own?

Let the time in hours to build the tree house alone be K and W, so K=W+3. The rate (house per hour) at which they can each build a house alone is 1/(W+3) and 1/W because hours per house is the inverse of house per hour. Working together their combined rate is 1/2 house per hour, because it takes 2 hours to build one house. So, combining the rates we get 1/(W+3)+1/W=1/2. We solve this by multiplying by the LCM 2W(W+3): 2W+2(W+3)=W(W+3); 4W+6=W^2+3W; W^2-W-6=0=(W-3)(W+2), so W=3 and K=6. SOLUTION: Kevin takes 6 hours to build the tree house by himself and Walter takes 3 hours. Or think of it like this. The house is made of 60 pieces of wood, say. After building the house together for an hour they would have used 30 pieces of wood, Walter would have used 20 pieces and Kevin 10. So it would have taken Walter 3 hours to make the house because 3*20=60, and Kevin would have taken 6 hours because 6*10=60. Working together 2(20+10)=60.

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