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Morgarella [4.7K]
3 years ago
9

There are 8 2/3 pounds of walnuts in a container which will be divided equally into containers that hold 1 1/5 pounds this would

fit _ and 4/18 containers
Mathematics
1 answer:
aev [14]3 years ago
7 0

we'll do the same as before, turning the mixed fractions to improper and do the division, keeping in mind that is simply asking how many times 1⅕ goes into 8⅔.


\bf \stackrel{mixed}{8\frac{2}{3}}\implies \cfrac{8\cdot 3+2}{3}\implies \stackrel{improper}{\cfrac{26}{3}}\\\\\\\stackrel{mixed}{1\frac{1}{5}}\implies \cfrac{1\cdot 5+1}{5}\implies \stackrel{improper}{\cfrac{6}{5}}\\\\[-0.35em]\rule{34em}{0.25pt}\\\\\cfrac{26}{3}\div \cfrac{6}{5}\implies \cfrac{26}{3}\cdot \cfrac{5}{6}\implies \cfrac{130}{18}\implies \cfrac{126+4}{18}\implies \cfrac{126}{18}+\cfrac{4}{18}\\\\\\\boxed{7+\cfrac{4}{18}}\implies 7\frac{4}{18}

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Step-by-step explanation:

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Are 12:14 and 60:70 equivalent ratios?<br><br> yes<br> no
NNADVOKAT [17]

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48% of their readers own a particular make of car. A marketing executive wants to test the claim that the percentage is actually
oksian1 [2.3K]

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There is no sufficient evidence to support the executive claim

Step-by-step explanation:

From the question we are told that

     The  population proportion is  p =  0.48

      The  sample proportion is  \r p  =  0.45

       The sample  size is  n  =  300

       The  level of significance is \alpha  =  0.02

The null hypothesis is  H_o  :  p=  0.48

The  alternative hypothesis is  H_a  :  p \ne  0.48

   Generally the test statistics is mathematically evaluated as

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=>        t =  \frac{0.45   -  0.48   }{ \sqrt{ \frac{0.48 (1 -  0.48 )}{300} } }

=>       t =  -1.04

The  p-value is mathematically represented as

     p-value  =  2P(z  >  |-1.04|)

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                 P(z  >  |-1.04|)  =   0.15

=>   p-value  =  2 * 0.15

=>   p-value  = 0.3

Given that  p-value  >  \alpha  we fail to reject the null hypothesis

   Hence we can conclude that there is no sufficient evidence to support the executive claim

5 0
3 years ago
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