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svp [43]
4 years ago
10

18/90 in simplest form

Mathematics
2 answers:
mafiozo [28]4 years ago
7 0
18/90
Divide by the common factor of 18 and 90, 9.
(18/9) / (90/9)
2/10
Divide top and bottom by 2

1/5

Hope this helps :)
Nitella [24]4 years ago
6 0
1/5 because  if you divide 18/90 by 2 it is 9/45 thn divide by 9 its 1/5
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You are given the following information obtained from a random sample of 5 observations. 20 18 17 22 18 At 90% confidence, you w
Margaret [11]

Answer:

a

  The null hypothesis is  

         H_o  : \mu  =  21

The Alternative  hypothesis is  

           H_a  :  \mu<   21

b

     \sigma_{\= x} =   0.8944

c

   t = -2.236

d

  Yes the  mean population is  significantly less than 21.

Step-by-step explanation:

From the question we are given

           a set of  data  

                               20  18  17  22  18

       The confidence level is 90%

       The  sample  size  is  n =  5  

Generally the mean of the sample  is  mathematically evaluated as

        \= x  =  \frac{20 + 18 +  17 +  22 +  18}{5}

       \= x  =  19

The standard deviation is evaluated as

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

         \sigma =  \sqrt{ \frac{ ( 20- 19 )^2 + ( 18- 19 )^2 +( 17- 19 )^2 +( 22- 19 )^2 +( 18- 19 )^2 }{5} }

         \sigma = 2

Now the confidence level is given as  90 %  hence the level of significance can be evaluated as

         \alpha = 100 - 90

        \alpha = 10%

         \alpha =0.10

Now the null hypothesis is  

         H_o  : \mu  =  21

the Alternative  hypothesis is  

           H_a  :  \mu<   21

The  standard error of mean is mathematically evaluated as

         \sigma_{\= x} =   \frac{\sigma}{ \sqrt{n} }

substituting values

         \sigma_{\= x} =   \frac{2}{ \sqrt{5 } }

        \sigma_{\= x} =   0.8944

The test statistic is  evaluated as  

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

substituting values

              t =  \frac{ 19  - 21 }{ 0.8944 }

              t = -2.236

The  critical value of the level of significance is  obtained from the critical value table for z values as  

                   z_{0.10} =  1.28

Looking at the obtained value we see that z_{0.10} is greater than the test statistics value so the null hypothesis is rejected

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Step-by-step explanation:

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