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tia_tia [17]
3 years ago
15

Your employee have expressed an interest in flextime, and would like to work 4 days a week instead of 5. The employees have been

working 9-hour days, including an hour for lunch. How many hours will they need to work daily if they switch to the new schedule
Mathematics
1 answer:
Flura [38]3 years ago
5 0

5 nine-hour days is a total of 5×9 = 45 hours.

If 45 hours are worked in 4 days, the daily average is

... 45/4 = 11 1/4 . . . hours

They will need to work <em>11 hours 15 minutes</em> daily if they switch.

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Answer: C

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7 0
3 years ago
In one day, The Tool Shed rented out 25 self-propelled and riding mowers for a total fee of $900. The dolly rental cost was $20
masha68 [24]

They rented out 5 self-propelled mowers and 20 riding mowers.

Data;

  • Total number of tool rented out = 25
  • Cost of tools rented = $900

<h3>System of Equations</h3>

To solve this problem, we have to write a system of equations to represent the problem.

since we have

  • x = self - propelled mower
  • y = riding mowers

Let's write equations with these variables.

x+y = 25...equation (i)\\20x+40y = 900...equation(ii)

From equation (i)

x+y= 25\\x = 25 - y...equation(iii)

Put equation (iii) into equation (ii)

20x+40y=900\\x = 25-y\\20(25-y)+40y=900\\\\500-20y+40y=900\\500+20y=900\\20y=900-500\\20y=400\\\frac{20y}{20}=\frac{400}{20}\\ y=20

Let's substitute the value of y into equation(i)

x+y =25\\x+20 = 25\\x = 25 - 20\\x = 5

From the calculations above, they rented 5 self-propelled mowers and 20 riding mowers.

Learn more on system of equations here;

brainly.com/question/13729904

4 0
2 years ago
Anyone please help me
o-na [289]

Answer: can’t help

Step-by-step explanation:

4 0
3 years ago
What's the answer to that equation above
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7 0
3 years ago
Determine which relation is a function. Question 13 options: a) {(3, 0), (– 2, – 2), (7, – 2), (– 2, 0)} b) c) y = 15x + 2 y = 1
antiseptic1488 [7]

Answer:

x=3−2d,5,−2(1+d),5,−27−2d,5,−2(2+d),5,2(y−d),5

Step-by-step explanation:Solving for x. Want to solve for y or solve for d instead?

1 Simplify  0-20−2  to  -2−2.

3,-2,-27,-2-2,02y=1,5x+2d3,−2,−27,−2−2,02y=1,5x+2d

2 Simplify  -2-2−2−2  to  -4−4.

3,-2,-27,-4,02y=1,5x+2d3,−2,−27,−4,02y=1,5x+2d

3 Subtract 2d2d from both sides.

3-2d,-2-2d,-27-2d,-4-2d,02y-2d=1,5x3−2d,−2−2d,−27−2d,−4−2d,02y−2d=1,5x

4 Divide both sides by 1,51,5.

\frac{3-2d}{1},5,\frac{-2-2d}{1},5,\frac{-27-2d}{1},5,\frac{-4-2d}{1},5,\frac{02y-2d}{1},5=x

​1

​

​3−2d

​​ ,5,

​1

​

​−2−2d

​​ ,5,

​1

​

​−27−2d

​​ ,5,

​1

​

​−4−2d

​​ ,5,

​1

​

​02y−2d

​​ ,5=x

5 Factor out the common term 22.

\frac{3-2d}{1},5,\frac{-2(1+d)}{1},5,\frac{-27-2d}{1},5,\frac{-4-2d}{1},5,\frac{02y-2d}{1},5=x

​1

​

​3−2d

​​ ,5,

​1

​

​−2(1+d)

​​ ,5,

​1

​

​−27−2d

​​ ,5,

​1

​

​−4−2d

​​ ,5,

​1

​

​02y−2d

​​ ,5=x

6 Factor out the common term 22.

\frac{3-2d}{1},5,\frac{-2(1+d)}{1},5,\frac{-27-2d}{1},5,\frac{-2(2+d)}{1},5,\frac{02y-2d}{1},5=x

​1

​

​3−2d

​​ ,5,

​1

​

​−2(1+d)

​​ ,5,

​1

​

​−27−2d

​​ ,5,

​1

​

​−2(2+d)

​​ ,5,

​1

​

​02y−2d

​​ ,5=x

7 Factor out the common term 22.

\frac{3-2d}{1},5,\frac{-2(1+d)}{1},5,\frac{-27-2d}{1},5,\frac{-2(2+d)}{1},5,\frac{2(y-d)}{1},5=x

​1

​

​3−2d

​​ ,5,

​1

​

​−2(1+d)

​​ ,5,

​1

​

​−27−2d

​​ ,5,

​1

​

​−2(2+d)

​​ ,5,

​1

​

​2(y−d)

​​ ,5=x

8 Simplify  \frac{3-2d}{1}

​1

​

​3−2d

​​   to  (3-2d)(3−2d).

3-2d,5,\frac{-2(1+d)}{1},5,\frac{-27-2d}{1},5,\frac{-2(2+d)}{1},5,\frac{2(y-d)}{1},5=x3−2d,5,

​1

​

​−2(1+d)

​​ ,5,

​1

​

​−27−2d

​​ ,5,

​1

​

​−2(2+d)

​​ ,5,

​1

​

​2(y−d)

​​ ,5=x

9 Simplify  \frac{-2(1+d)}{1}

​1

​

​−2(1+d)

​​   to  (-2(1+d))(−2(1+d)).

3-2d,5,-2(1+d),5,\frac{-27-2d}{1},5,\frac{-2(2+d)}{1},5,\frac{2(y-d)}{1},5=x3−2d,5,−2(1+d),5,

​1

​

​−27−2d

​​ ,5,

​1

​

​−2(2+d)

​​ ,5,

​1

​

​2(y−d)

​​ ,5=x

10 Simplify  \frac{-27-2d}{1}

​1

​

​−27−2d

​​   to  (-27-2d)(−27−2d).

3-2d,5,-2(1+d),5,-27-2d,5,\frac{-2(2+d)}{1},5,\frac{2(y-d)}{1},5=x3−2d,5,−2(1+d),5,−27−2d,5,

​1

​

​−2(2+d)

​​ ,5,

​1

​

​2(y−d)

​​ ,5=x

11 Simplify  \frac{-2(2+d)}{1}

​1

​

​−2(2+d)

​​   to  (-2(2+d))(−2(2+d)).

3-2d,5,-2(1+d),5,-27-2d,5,-2(2+d),5,\frac{2(y-d)}{1},5=x3−2d,5,−2(1+d),5,−27−2d,5,−2(2+d),5,

​1

​

​2(y−d)

​​ ,5=x

12 Simplify  \frac{2(y-d)}{1}

​1

​

​2(y−d)

​​   to  (2(y-d))(2(y−d)).

3-2d,5,-2(1+d),5,-27-2d,5,-2(2+d),5,2(y-d),5=x3−2d,5,−2(1+d),5,−27−2d,5,−2(2+d),5,2(y−d),5=x

13 Switch sides.

x=3-2d,5,-2(1+d),5,-27-2d,5,-2(2+d),5,2(y-d),5x=3−2d,5,−2(1+d),5,−27−2d,5,−2(2+d),5,2(y−d),5

Done

5 0
3 years ago
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