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Fed [463]
2 years ago
9

Someone please help asap! I will mark as brainliest, thank you!!

Mathematics
1 answer:
madreJ [45]2 years ago
6 0

Answer:

19 square units

Step-by-step explanation:

Hello!

Step 1: Plot the points

We are first going to plot the points on the coordinate graph and connect the points to create a shape.

Refer to attachment titled "Step 1"

Step 2: Box the shape

Since this is an irregular shape, I'm going to make a box around the shape and subtract the area of the empty spaces to find the area of the original polygon.

Refer to attachment titled "Step 2"

Step 3: Area of Edges

We are going to find the area of the shapes A, B, and C and subtract them from the area of the big box to find the area of the polygon.

Area of a triangle = 1/2(base x height)

  • Area of Box = 6 un x 7 un
  • Area of Box = 42 un²

Area of A:

  • Area A = 1/2(4 un x 2 un)
  • Area A = 1/2(8 un²)
  • Area A = 4 un²

Area of B:

  • Area B = 4 un x 4 un
  • Area B = 16 un²

Area of C:

  • Area C = 1/2(2 un x 3 un)
  • Area C = 1/2(6 un²)
  • Area C = 3 un²

Step 4: Area of the polygon

Now, we can simply subtract the values of A, B, and C from the box to get the area of the polygon.

  • Area Polygon = Box - A - B - C
  • Area Polygon = 42 un² - 4 un² - 16 un² - 3 un²
  • Area Polygon = 38 un² - 16 un² - 3 un²
  • Area Polygon = 22 un² - 3 un²
  • Area Polygon = 19 un²

The area of the polygon is 19 un²

-Chetan K

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Answer:

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Step-by-step explanation:

Let assume that the number of computer produced by factory C is  k = 1  

 So  From the  question we are told that

       The number produced by  factory A is  4k =  4

        The  number produced by factory B is  7k  = 7

        The  probability of defective computers from A is  P(A) =  0.04

        The  probability of defective computers from B is  P(B)  =  0.02

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Now the probability of factory A producing a defective computer out of the 4 computers produced is  

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substituting values

        P(a) =  4 * 0.04

        P(a) = 0.16

The probability of factory B producing a defective computer out of the 7 computers produced is  

       P(b) = 7  *  P(B)

substituting values

        P(b) =  7 * 0.02

        P(b) = 0.14

The probability of factory C producing a defective computer out of the 1 computer produced is  

       P(c) = 1  *  P(C)

substituting values

        P(c) =  1 * 0.03

        P(b) = 0.03

So the probability that the a computer produced from the three factory will be defective is  

     P(t) =  P(a) +  P(b) +  P(c)

substituting values

     P(t) =   0.16  + 0.14 +  0.03

     P(t) =   0.33

Now the probability that the defective computer is produced from factory A is

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       P(A') =  \frac{ 0.16}{0.33}

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