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3 years ago

PLEASE HELP

1 answer:

7 0

If the equation for volume is V=l*w*h, you can plug in all of the information you know. When you do this, the equation becomes 1440=15*12*h, or 1440=180*h (because 15*12=180). You can divide both sides by 180 to get h. 1440/180=8, do the depth of the pool is 8 ft. You can also check this answer by making sure that 15*12*8 is equal to 1440 (which it is). I hope this helped!

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Explain how the following are related: linear relationship, proportional relationship, equivalent ratios, and direct variation

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With all of the following you are able to make a graph, table, and a equation.

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2 years ago

A property owner wants to build a cyclone fence around their property. The lot is rectangular, 110 feet by 270 feet. The fence i

dimulka [17.4K]

Answer:

total cost= $5,814

Step-by-step explanation:

Total linear feet measured: (110 x 2) + (270 x 2) = 760 linear ft

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3 years ago

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Answer: 24

Step-by-step explanation:

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3 0

1 year ago

Read 2 more answers

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Answer:

x>-4

Step-by-step explanation:

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3 0

3 years ago

Wal-Mart conducted a study to check the accuracy of checkout scanners at its stores. At each of the 60 randomly selected Wal-Mar

Mamont248 [21]

Answer:

The 95% confidence interval is

$0.7811 < p < 0.9529$

Generally the interval above can interpreted as

There is 95% confidence that the true proportion of Wal-Mart stores that have more than 2 items priced inaccurately per 100 items scanned lie within the interval

Generally 99% is outside the interval obtained in a above then the claim of Wal-mart is not believable

$n = 125 \ stores$

Step-by-step explanation:

From the question we are told that

The sample size is n = 60

The number of stores that had more than 2 items price incorrectly is k = 52

Generally the sample proportion is mathematically represented as

$\^ p = \frac{ k }{ n }$

=> $\^ p = \frac{ 52 }{ 60 }$

=> $\^ p = 0.867$

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 normal distribution table the critical value of $\frac{\alpha }{2}$ is

$Z_{\frac{\alpha }{2} } = 1.96$

Generally the margin of error is mathematically represented as

$E = Z_{\frac{\alpha }{2} } * \sqrt{\frac{\^ p (1- \^ p)}{n} }$

=> $E = 1.96 * \sqrt{\frac{ 0.867 (1- 0.867)}{60} }$

=> $E = 0.0859$

Generally 95% confidence interval is mathematically represented as

$\^ p -E < p < \^ p +E$

=> $0.867 - 0.0859 < p < 0.867 + 0.0859$

=> $0.7811 < p < 0.9529$

Generally the interval above can interpreted as

There is 95% confidence that the true proportion of Wal-Mart stores that have more than 2 items priced inaccurately per 100 items scanned lie within the interval

Considering question b

Generally 99% is outside the interval obtained in a above then the claim of Wal-mart is not believable

Considering question c

From the question we are told that

The margin of error is E = 0.05

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 normal distribution table the critical value of $\frac{\alpha }{2}$ is

$Z_{\frac{\alpha }{2} } = 1.645$

Generally the sample size is mathematically represented as

$n = [\frac{Z_{\frac{\alpha }{2} }}{E} ]^2 * \^ p (1 - \^ p )$

=> $n= [\frac{1.645 }}{0.05} ]^2 * 0.867 (1 - 0.867 )$

=> $n = 125 \ stores$

4 0

3 years ago