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Tcecarenko [31]
3 years ago
11

Write two polynomials with a difference of 2x2 + x + 3

Mathematics
1 answer:
Anettt [7]3 years ago
5 0

Answer:

The two polynomial CAN be 5x^4+5x^3+8x^2+x+6 and 5x^4+5x^3+8x^2+2x+13.

Step-by-step explanation:

Step 1: <u>Simplify the expression:</u>

2*2 + x + 3

4+x+3

x+7

Step 2: <u>Think of a random polynomial:</u>

5x^4+5x^3+8x^2+x+6

Step 3: <u>Either add your "random polynomial" by </u>x+7<u> or subtract it.</u>

<u>I want to add it so...</u>

5x^4+5x^3+8x^2+x+6+x+7

= 5x^4+5x^3+8x^2+2x+13

 Check by subtracting  5x^4+5x^3+8x^2+x+6 and 5x^4+5x^3+8x^2+2x+13.

         5x^4+5x^3+8x^2+2x+13

   -      5x^4+5x^3+8x^2+x+6

        _____________________

         0x^4 - 0x^3 -0x^2 +x +7 <em><u>or  </u></em>x+7<em><u /></em>

         

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Step-by-step explanation:

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Six-four million, one hundred eighty-six thousand, three hundred square miles in standard form
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Answer:

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I'm sorry if I misunderstood.

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The probability that the sample proportion is within ± 0.02 of the population proportion is 0.3328

<h3>How to determine the probability?</h3>

The given parameters are:

  • Sample size, n = 100
  • Population proportion, p = 82%

Start by calculating the mean:

\mu = np

\mu = 100 * 82\%

\mu = 82

Calculate the standard deviation:

\sigma = \sqrt{\mu(1 - p)

\sigma = \sqrt{82 * (1 - 82\%)

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Within ± 0.02 of the population proportion are:

x_{min} = 82 * (1 - 0.02) = 80.38

x_{max} = 82 * (1 + 0.02) = 83.64

Calculate the z-scores at these points using:

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So, we have:

z_1 = \frac{80.36 - 82}{3.84} = -0.43

z_2 = \frac{83.64 - 82}{3.84} = 0.43

The probability is then represented as:

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Using the z table of probabilities, we have:

P(x ± 0.02) = 0.3328

Hence, the probability that the sample proportion is within ± 0.02 of the population proportion is 0.3328

Read more about probability at:

brainly.com/question/25870256

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