Can you please help me

The number of tissues that can fit the container will be given by:
[volume of the container]/[volume of tissue box]
volume of container:
V=length×width×height
V=19.5×39×19.5
V=14829.75

Volume of tissue box:
v=6.5×6.5×6.5
v=274.625

Number of tissues in the container:
14829.75/274.625
=54

Answer: 54 tissue paper boxes

3 0

3 years ago

Use the given information to find the p-value. Also, use a 0.05 significance level and state the conclusion about the null hyp

Hatshy [7]

Answer:

We can assume that the statistic is $z_{calc}=0.78$

$p_v = 2* P(z>0.78) = 0.435$

So the p value obtained was a very high value and using the significance level given $\alpha=0.05$ we have $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the true proportion of interest is not different from 3/5

Step-by-step explanation:

Concepts and formulas to use

We need to conduct a hypothesis in order to test the claim that the true proportion is equal to 3/5 or not.:

Null hypothesis: $p=3/5$

Alternative hypothesis: $p \neq 3/5$

When we conduct a proportion test we need to use the z statistic, and the is given by:

$z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}$ (1)

The One-Sample Proportion Test is used to assess whether a population proportion $\hat p$ is significantly different from a hypothesized value $p_o$ .

Calculate the statistic

We can assume that the statistic is $z_{calc}=0.78$

Statistical decision

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.

The significance level provided $\alpha=0.05$ . The next step would be calculate the p value for this test.

Since is a bilateral test the p value would be:

$p_v = 2* P(z>0.78) = 0.435$

So the p value obtained was a very high value and using the significance level given $\alpha=0.05$ we have $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the true proportion of interest is not different from 3/5

7 0

3 years ago

BRAINLIEST ASAP! PLEASE HELP ME :)

Mariana [72]

Answer:

Yes, the given measures can be the lengths of the sides of a triangle.

Step-by-step explanation:

hope this helps

5 0

3 years ago

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Can you please help me​

Can you please help me