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Illusion [34]
3 years ago
15

What is the are of the figure below​

Mathematics
2 answers:
qwelly [4]3 years ago
7 0
The answer is 8 cm squared
Vladimir [108]3 years ago
4 0

Answer:

The answer would most likely to eight. If not use the formula Base x Height

Step-by-step explanation:

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3 years ago
The areas of 10 cities are given in the table.
Masteriza [31]
The range for the cities in the south is 145 larger than the range for the cities in the north.

We find the range by subtracting the highest and lowest values; in the North, we have:
305 - 58 = 247

In the South, we have:
503 - 111 = 392

The difference in the ranges is 392 - 247 = 145.
5 0
3 years ago
Read 2 more answers
Please help me solve this
Afina-wow [57]

Answer:

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Step-by-step explanation:

8 0
3 years ago
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statuscvo [17]

Answer:

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Step-by-step explanation:

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5 0
2 years ago
A randomized trial tested the effectiveness of diets on adults. Among 36 subjects using Diet 1, the mean weight loss after a yea
seropon [69]

Answer:

The 95%  confidence interval is

           0.45 <  \mu_1 - \mu_2  < 5.35

Step-by-step explanation:

From the question we are told that

   The first sample size is  n_1   =  36

   The first  sample mean is  \= x_1  =  3.5

   The first standard deviation is  \sigma_1  =  5.9 \ pounds

   The second  sample size is n_2 =  36

    The second  sample mean is  \= x_2 =  0.6

    The second  standard deviation is \sigma  =  4.4

Generally the degree of freedom is mathematically represented as

     df =  \frac{ [ \frac{s_1^2 }{n_1 }  + \frac{s_2^2 }{n_2} ]^2 }{ \frac{1}{(n_1 - 1 )} [ \frac{s_1^2}{n_1} ]^2 + \frac{1}{(n_2 - 1 )} [ \frac{s_2^2}{n_2} ]^2  }

=>  df =  \frac{ [ \frac{5.9^2 }{34 }  + \frac{4.4^2 }{34} ]^2 }{ \frac{1}{(34 - 1 )} [ \frac{5.9^2}{34} ]^2 + \frac{1}{(34- 1 )} [ \frac{4.4^2}{ 34} ]^2  }

=>  df =63

Generally the standard error is mathematically represented as

      SE =  \sqrt{ \frac{s_1 ^2 }{n_1}  + \frac{s_2^2 }{ n_2 } }

=>  SE =  \sqrt{ \frac{ 5.9 ^2 }{ 36 }  + \frac{ 4.4^2 }{36} }

=>  SE = 1.227

From the question we are told the confidence level is  95% , hence the level of significance is    

      \alpha = (100 - 95 ) \%

=>   \alpha = 0.05

Generally from the t distribution table the critical value  of   at a degree of freedom of is  

     t_{\frac{\alpha }{2}, 63  } =  1.998

Generally the margin of error is mathematically represented as

        E =  t_{\frac{\alpha }{2}, 63  } *  SE

=>    E =  1.998 * 1.227

=>    E =  2.45

Generally 95% confidence interval is mathematically represented as  

      (\= x_1 - \x_2) -E <  \mu

 => (3.5  - 0.6) - 2.45 <  \mu_1 - \mu_2  < (  3.5  - 0.6)  + 2.45

=>   0.45 <  \mu_1 - \mu_2  < 5.35

     

8 0
2 years ago
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