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s2008m [1.1K]
2 years ago
12

Find the second derivative y=1/5x^2+1/11x

Mathematics
1 answer:
Amiraneli [1.4K]2 years ago
8 0

\frac{dy}{dx}=\frac{1}{5}.x^{2-1}.2 + \frac{1}{11}.x^{1-1}.1

\frac{dy}{dx}=\frac{2}{5}.x+\frac{1}{11}

\frac{d^2y}{dx^2}=\frac{2}{5}.x^{1-1}.1

\frac{d^2y}{dx^2}=\frac{2}{5}

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A random sample of 300 circuits generated 13 defectives. Use the data to test the hypothesis Upper H Subscript 0 Baseline colon
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Complete Question

A random sample of 300 circuits generated 13 defectives. a. Use the data to test

                            H_o : p = 0.05

Versus

 

                          H_1 : p \ne 0.05

Use α = 0.05. Find the P-value for the test

   

Answer:

The  p-value is  p-value =  0.5949        

Step-by-step explanation:

From the question we are told that

  The sample size is  n = 300

    The number of defective circuits is  k = 13

Generally the sample proportion of defective circuits is mathematically represented as

        \^ p = \frac{k}{n}

=>     \^ p = \frac{13}{300}

=>     \^ p = 0.0433

Generally the standard Error is mathematically represented as

       SE  = \sqrt{\frac{p(1- p)}{n} }

=>     SE  = \sqrt{\frac{0.05(1- 0.05)}{300} }

=>     SE  = 0.0126

Generally the test statistics is mathematically represented as

       z =  \frac{\^ p - p }{SE}

=>     z =  \frac{0.0433 - 0.05 }{0.0126}

=>     z =  -0.5317

From the z table  the area under the normal curve to the left  corresponding to  -0.5317  is

         (P < -0.5317 ) = 0.29747

Generally the p-value is mathematically represented as

       p-value =  2 * P(Z <  -0.5317 )

=>     p-value =  2 * 0.29747

=>     p-value =  0.5949        

     

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3 years ago
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