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Vesnalui [34]
3 years ago
13

the histogram on the left shows the number of hours students in an art class practiced drawing during one week.​

Mathematics
1 answer:
svet-max [94.6K]3 years ago
4 0

Step-by-step explanation:

mmm Next no se...........

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Please help solve questions
Alekssandra [29.7K]

Take some points

  • 2x+3y<6
  • 3y<-2x+6
  • y<-2/3x+2

As here < sign present line will be dashed and shading should be done below the line

So

  • (1,1)
  • (1,0)
  • (2,0)
  • (-1,2)

Graph attached

4 0
2 years ago
Read 2 more answers
What is the slope-intercept form of the function described by this table?
Vikentia [17]
Y=5x - 3 Because the difference in the y values are 5, the rise is 5. That is the TOP of the slope. The difference in the x values is 1, that is the run or the BOTTOM of the slope. So m = 5/1. Using y = mx + b 2 = 5(1) + b -3 = b Therefore, y = 5x -3
8 0
3 years ago
Which of the following expressions is equal to 1 divided by 16
Ber [7]
The answer is 1/2 4
3 0
2 years ago
Eleven more than two
Veronika [31]

Answer:13

Step-by-step explanation:

11+2=13

6 0
3 years ago
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
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