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pogonyaev
3 years ago
13

Graph. y-2=2/3(x+4).

Mathematics
1 answer:
lara31 [8.8K]3 years ago
6 0

Answer:

M = 2/3

Y-Intercept = 14/3


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Answer:

As x decreases in value.f(x) decreases in value.......

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The corporate team building event will cost $12 if it has three attendees if there are five attendees how much was a quaker team
Norma-Jean [14]

Answer:

Cost to arrange the event for 5 attendees will be $20.

Step-by-step explanation:

We have to solve this question by calculating the unit rates or unitary method.

Since cost to arrange corporate team building event for 3 attendees is = $12

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Therefore, cost to arrange the event for five attendees will be = 5×4 = $20

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yawa3891 [41]

Answer:

This is not a statistical question

Step-by-step explanation:

A statistical question is a question the gathers varying data

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Santa Claus is assigning elves to work an eight-hour shift making toy trucks. Apprentice earn five candy canes per hour but can
dimaraw [331]

Answer:

  (a) 5 senior, 4 apprentice

  (b) 368 per shift

  (c) 7.5 senior, 0 apprentice

Step-by-step explanation:

The problem can be described by two inequalities. On describes the limit on the number of elves in the shop; the other describes the limit on the total payroll. Let x and y represent the number of apprentice and senior elves, respectively. Then the inequalities for the first scenario are ...

  x + y ≤ 9 . . . . . . . . . . . . total number of elves in the shop

  5x +8y ≤ 480/8 . . . . . . candy canes per hour paid to elves

These two inequalities are graphed in the first attachment. They describe a solution space with vertices at ...

  (x, y) = (0, 7.5), (4, 5), (9, 0)

__

(a) Santa wants to  maximize the output of trucks, so wants to maximize the function t = 4x +6y.

At the vertices of the solution space, the values of this function are ...

  t(0, 7.5) = 45

  t(4, 5) = 46

  t(9, 0) = 36

Output of trucks is maximized by a workforce of 4 apprentice elves and 5 senior elves.

__

(b) The above calculations show 46 trucks per hour can be made, so ...

  46×8 = 368 . . . trucks in an 8-hour shift

__

(c) The new demands change the inequalities to ...

  x + y ≤ 8 . . . . . . number of workers

  7x +8y ≤ 60 . . . total wages (per hour)

The vertices of the feasible region for these condtions are ...

  (x, y) = (0, 7.5), (4, 4), (8, 0)

From above, we know the truck output will be maximized at the vertex (x, y) = (0, 7.5). However, we know we cannot have 7.5 senior elves working in the shop. We can have 7 or 8 elves working.

If the workforce must remain constant, truck output is maximized by a workforce of 7 senior elves.

If the workforce can vary through the shift, truck output is maximized by adding one more senior elf in the shop for half a shift.

Santa should assign 7 senior elves for the entire shift, and 8 senior elves (one more) for half a shift.

_____

<em>Comment on apprentice elf wages</em>

At 5 candy canes for 4 trucks, apprentice elves produced trucks for a cost of 1.25 candy canes per truck. At 8 candy canes for 6 trucks, senior elves produced trucks for a cost of about 1.33 candy canes per truck. The reason for employing senior elves in the first scenario is that their productivity is 1.5 times that of apprentice elves while their cost per truck is about 1.07 times that of apprentice elves.

After the apprentice elves wages were increased, their cost per truck is 1.75 candy canes per truck, but their productivity hasn't changed. They have essentially priced themselves out of a job, because they are not competitive with senior elves.

5 0
2 years ago
3.
Ostrovityanka [42]

Answer:

110÷360×π×4.6^(2)= 20.312

Step-by-step explanation:

Area of sector = angle / 360 x π x r^(2)

8 0
3 years ago
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