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Softa [21]
3 years ago
8

A mixed of purple paint contains 4 teaspoons of red paint and 10 teaspoons of blue paint.To make the same shade of purple paint

using 40 teaspoons of blue paint how much red would you need? A.12 B.14 C.16 D.160
Mathematics
1 answer:
lina2011 [118]3 years ago
5 0

Answer:

Number of  teaspoon of red paint = 16

Step-by-step explanation:

Given:

Mixture contain = 4 teaspoon red paint and 10 teaspoons blue paint

New Mixture = 40 teaspoons blue paint

Find:

Number of  teaspoon of red paint

Computation:

4 teaspoon red paint / 10 teaspoons blue paint = Number of  teaspoon of red paint / 40 teaspoons blue paint

Number of  teaspoon of red paint = 16

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Step-by-step explanation:

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Answer:

a

The null hypothesis is  H_o : \mu  =  13.4000

The alternative hypothesis is  H_a :  \mu \ne  13.4000

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b

The  99% confidence level is   13.3930  < \mu  < 13.3994

Step-by-step explanation:

From the question we are told that

  The sample size is  n =  10

   The  population mean is  \mu =  13.4 000 \  angstroms

   The level of significance is  \alpha =  0.05

   The  sample data is  

13.3987, 13.3957, 13.3902, 13.4015, 13.4001, 13.3918, 13.3965, 13.3925, 13.3946, and 13.4002

Generally the sample mean is mathematically represented as

        \= x =  \frac{13.3987+ 13.3957\cdots +13.4002 }{10}

=>     \= x = 13.3962

Generally the sample standard deviation  is mathematically represented as

    \sigma = \sqrt{\frac{\sum (x_i - \= x)^2}{n} }

=> \sigma = \sqrt{\frac{ (13.3987 - 13.3962)^2 +  (13.3987 - 13.3962)^2 + \cdots + (13.3987 -13.4002)^2  }{10} }

=>  \sigma =0.0039

The null hypothesis is  H_o : \mu  =  13.4000

The alternative hypothesis is  H_a :  \mu \ne  13.4000

Generally the test statistics is mathematically represented as

      z  =  \frac{\= x - \mu}{\frac{\sigma}{\sqrt{n} } }

=>    z  =  \frac{13.3962 -  13.4000}{\frac{0.0039}{\sqrt{10} } }

=>   z =  3.08

Generally the p-value is mathematically represented as

       p-value  = 2P( z   >  3.08)

From the z-table  P(z >  3.08)= 0.001035

So

       p-value  = 2* 0.001035

      p-value  = 0.00207

So from the obtained value we see that

     p-value  < \alpha

Hence the null hypothesis is rejected

Consider the b question

Given that the confidence level is  99%  then the level of significance is

    \alpha =  (100 -99)\%

=> \alpha =  0.01

Generally from the normal distribution table critical value  of  \frac{\alpha }{2} is  

    Z_{\frac{\alpha }{2} } =  2.58

Generally the margin of error is mathematically represented as  

     E =  Z_{\frac{\alpha }{2} } *  \frac{\sigma}{\sqrt{n} }

=>   E = 2.58  *  \frac{0.0039}{\sqrt{10} }

=>    E = 0.00318

Generally the 99% confidence interval is mathematically represented as

     13.3962   - 0.00318  < \mu  < 13.3962   +  0.00318

=>   13.3930  < \mu  < 13.3994

 

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