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pantera1 [17]
3 years ago
13

A truck is carrying 10 3 bushels of apples, 10 2 bushels of grapes, and 10 1 bushels of oranges. Each bushel of apples weighs

32 pounds, each bushel of grapes weighs 25 pounds, and each bushel of oranges weighs 42 pounds. What is the total weight of the fruit in pounds?
Mathematics
1 answer:
almond37 [142]3 years ago
5 0

Answer:

10,088 pounds

Step-by-step explanation:

Given

103 bushels of apple

102 bushels of grape

101 bushels of oranges

Also weight of each bushel:

Apple = 32 pounds

Grape = 25 pounds

Oranges = 42 pounds

Total weight will be sum of product of each:

Total Weight: 103(32) + 102(25) + 101(42) = 10,088 pounds

Amara Mcnair
2 years ago
Thanks
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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

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              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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Answer:

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Step-by-step explanation:

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