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Arisa [49]
3 years ago
9

A baseball bat factory produces 24,000 bats per day and uses 50 boxes for packing the bats. If the number of boxes needed for pa

cking is proportional to the number of bats produced, how many boxes would be needed on a day when 36,000 bats are produced?
Mathematics
2 answers:
Anni [7]3 years ago
7 0

Answer:75

Step-by-step explanation:

Bats 24000/50

Box 480

Kisachek [45]3 years ago
3 0

Answer:

Total boxes= 75

Step-by-step explanation:

Giving the following information:

A baseball bat factory produces 24,000 bats per day and uses 50 boxes for packing the bats.

<u>First, we need to determine the number of bats that fits into a box:</u>

Bats per box= 24,000 / 50

Barts per box= 480

<u>Now, the number of boxes for 36,000 bats:</u>

Total boxes= total bats / bats per box

Total boxes= 36,000/480

Total boxes= 75

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Step-by-step explanation:

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6. Find the possible values of n in the inequality –3n &lt; 81. A. n &gt; –27 B. n &lt; 27 C. n = –27 D. n = 27
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State the linear programming problem in mathematical terms, identifying the objective function and the constraints. A firm makes
Sedbober [7]

Answer:

Maximum profit at (3,0) is $27.

Step-by-step explanation:

Let  quantity of  products A=x

Quantity  of products B=y

Product A takes time on machine L=2 hours

Product A takes time on machine M=2 hours

Product B takes time on machineL= 4 hours

Product B takes time on machine M=3 hours

Machine L can used total time= 8hours

Machine M can used total time= 6hours

Profit on product A= $9

Profit on product B=$7

According to question

Objective function Z=9x+7y

Constraints:

2x+3y\leq 6

2x+4y\leq 8

Where x\geq 0, y\geq 0

I equation 2x+3y\leq 6

I equation in inequality change into equality we get

2x+3y=6

Put x=0 then we get

y=2

If we put y=0 then we get

x= 3

Therefore , we get two points A (0,2) and B (3,0) and plot the graph for equation I

Now put x=0 and y=0 in I equation in inequality

Then we get 0\leq 6

Hence, this equation is true then shaded regoin is  below the line .

Similarly , for II equation

First change inequality equation into equality equation

we get 2x+4y=8

Put x= 0 then we get

y=2

Put y=0 Then we get

x=4

Therefore, we get two points C(0,2)a nd D(4,0) and plot the graph for equation II

Point  A and C are same

Put x=0 and y=0 in the in inequality equation II then we get

0\leq 8

Hence, this equation is true .Therefore, the shaded region is below the line.

By graph we can see both line intersect at the points A(0,2)

The feasible region is AOBA and bounded.

To find the value of objective function on points

A (0,2), O(0,0) and B(3,0)

Put A(0,2)

Z= 9\times 0+7\times 2=14

At point O(0,0)

Z=0

At point B(3,0)

Z=9\times3+7\times0=27

Hence maximum value of z= 27 at point B(3,0)

Therefore, the maximum profit is $27.

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Jobisdone [24]
The correct equation is y=1/2x-21
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