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3 years ago

15

a retangular pool is filled with 18.75 cubic meters of water. If the pool is 5 meters long and 2.5 meters wide, what is the dept

h of the pool

2 answers:

BigorU [14]3 years ago

7 0

The volume of a rectangular prism is given by length x width x height.

Now in the question the volume is given, length is given, and width is given.

horsena [70]3 years ago

6 0

Volume=length width x depth
18.75=5x2.5x depth
18.75=12.5x depth
Depth=1.5

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2 years ago

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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

$(\= x_1 - \x_2) -E < \mu$

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

=> $0.45 < \mu_1 - \mu_2 < 5.35$

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2 years ago