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Alina [70]
3 years ago
9

Which represents the solutions of the graphed system of equations, y=x^2-2x and y=-2x-1

Mathematics
1 answer:
pickupchik [31]3 years ago
7 0

Answer:

No real roots.  Imaginary roots would be x = plus or minus i .

Step-by-step explanation:

Eliminate y for now by equating y=x^2-2x and y=-2x-1.  Then:

x^2 - 2x = -2x - 1

Add 2x to both sides.  Then x^2 = -1.

This is NO SOLUTION, since the square root of -1 would be imaginary.

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Step-by-step explanation:

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Read 2 more answers
Silicon wafers are scored and then broken into the many small microchips that will he mounted into circuits. Two breaking method
antoniya [11.8K]

Answer:

a

The estimate is  - 0.0265\le  K \le  0.0465

b

Method B this is because the faulty breaks are less

Step-by-step explanation:

The number of microchips broken in method A is  n_1 = 400

The number of faulty breaks of method A is  X_1 = 32

 The number of microchips broken in method B is  n_2  = 400

 The number of faulty breaks of method A is  X_2 = 32

  The proportion of the faulty breaks to the total breaks in method A is

       p_1 = \frac{32}{400}

      p_1 = 0.08

 The proportion of the faulty to the total breaks in method B is

      p_2 =  \frac{28}{400}

     p_2 =  0.07

For this estimation the standard error is  

      SE =  \sqrt{ \frac{p_1 (1 - p_1)}{n_1}  + \frac{p_2 (1- p_2 )}{n_2} } }

  substituting values

       SE =  \sqrt{ \frac{0.08 (1 - 0.08)}{400}  + \frac{0.07 (1- 0.07 )}{400} } }

      SE = 0.0186

The z-values of confidence coefficient of 0.95 from the z-table is  

       z_{0.95} =  1.96

The difference between proportions of improperly broken microchips for the two breaking methods is mathematically represented as

        K = [p_1 - p_2 ] \pm z_{0.95} * SE

substituting values

        K = [0.08 - 0.07 ] \pm 1.96 *0.0186

         K  =  - 0.0265 \ or  \ K  =  0.0465

The interval of the difference between proportions of improperly broken microchips for the two breaking methods is  

      - 0.0265\le  K \le  0.0465

3 0
3 years ago
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