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algol [13]
3 years ago
15

Please help me with this!!

Mathematics
1 answer:
Naddik [55]3 years ago
3 0

Step-by-step explanation:

To find the y-coordinate points we need to evaluate the function for all the x values in the table. In other words, we need to replace x with each value in our given function and simplify.

- For x = 0

f(x)=(x-2)^2-5

f(0)=(0-2)^2-5

f(0)=(-2)^2-5

f(0)=4-5

f(x)=-1

Since x=0 and y=-1, our first point is (0, -1)

- For x = 1

f(x)=(x-2)^2-5

f(1)=(1-2)^2-5

f(1)=(-1)^2-5

f(1)=1-5

f(x)=-4

Our second point is (1, -4)

- For x = 2

f(x)=(x-2)^2-5

f(2)=(2-2)^2-5

f(2)=(0)^2-5

f(x)=-5

Our third point is (2, -5)

- For x = 3

f(x)=(x-2)^2-5

f(3)=(3-2)^2-5

f(3)=(1)^2-5

f(3)=1-5

f(x)=-4

Our fourth point is (3, -4)

- For x = 4

f(x)=(x-2)^2-5

f(4)=(4-2)^2-5

f(4)=(2)^2-5

f(4)=4-5

f(x)=-1

Our fifth point is (4, -1)

Now we just need to plot each point in our coordinate plane and join them with the parabola as you can see in the attached picture.

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A machine makes two components; call them types A and B. It takes 250 seconds to switch production between the component types.
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Step-by-step explanation:

The question is incomplete.

Question complete:

<em>"A machine makes two components, call them type A and B. It takes 250 seconds to switch production between the component types. During that time no production occurs. When in production, each unit of A or B requires 0.5 seconds to be completed. The two components, A and B are combined in an assembly process to make a final product, call it C. The assembly step can combine the two components into 1 unit every 2 seconds. What is the average inventory of B components?"</em>

We have two steps in this process: the manufacturing of A and B and the assembly process.

We are not told about the batch size, so we will treat that as a variable called n. This represents the amount of units (of A or B) that are manufactured per cycle.

Time (seconds) to manufacture n units of A and n units of B:

T_m=n*0.5+n*0.5+250=n+250

Time (seconds)to assemble n units of C

T_a=n*2=2n

Then the total time of the cycle is:

T_m+T_a=n+250+2n=3n+250

The average inventory of B is n units per cycle, which lasts (3n+250) seconds.

Inventory=\frac{n}{3n+250}

For a batch size of 100 units, the average inventory is:

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For a batch size of 1,000 units, the average inventory is:

I_{1000}=\frac{1000}{3*1000+250}=\frac{1000}{3250}=  0.308\, units/second

5 0
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