Need number 12 plz help

Solve the inequality for the missing variable<br> 24 -1t<br> 1/5

Scorpion4ik [409]

Answer:

−t+24

Step-by-step explanation:

8 0

3 years ago

which of the following fractions are equivalent? 3/4 and 6/7. 2/3 and 12/13. 1/9 and 4/12. 1/6 and 3/18

sertanlavr [38]

Answer:

1/6 and 3/18

Step-by-step explanation:

I used the butterfly method to compare all the fraction pairs together and then 1/6 and 3/18 were equal. Hopefully I can help you.

8 0

3 years ago

Lin’s goal is to run 5 miles. She ran 1⁄2 mile. What fraction of her goal is that?

lisov135 [29]

Answer:

2 1/2

Step-by-step explanation:

5 divided by 2

8 0

3 years ago

Read 2 more answers

Cameron is painting one wall of his dining room. It is

Harman [31]

Answer:

me name is cameron

Step-by-step explanation:

6 0

2 years ago

An e-commerce research company claims that 60% or more graduate students have bought merchandise on-line at their site. A consum

Elena-2011 [213]

Answer:

$z=\frac{0.55 -0.6}{\sqrt{\frac{0.6(1-0.6)}{80}}}=-0.913$

$p_v =P(z$

So the p value obtained was a very high value and using the significance level given $\alpha=0.05$ we see that $p_v>\alpha$ 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 proportion of graduate students show that only 44 students have ever done so is not significantly lower than 0.6

Step-by-step explanation:

Data given and notation

n=80 represent the random sample taken

X=44 represent the students that have bought merchandise on-line at their site

$\hat p=\frac{44}{80}=0.55$ estimated proportion of graduate students show that only 44 students have ever done so

$p_o=0.6$ is the value that we want to test

$\alpha$ represent the significance level

z would represent the statistic (variable of interest)

$p_v$ represent the p value (variable of interest)

Concepts and formulas to use

We need to conduct a hypothesis in order to test the claim that the true proportion of interest is lower than 0.6 or 60%, so then the system of hypothesis are.:

Null hypothesis: $p \geq 0.6$

Alternative hypothesis: $p < 0.6$

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

Since we have all the info requires we can replace in formula (1) like this:

$z=\frac{0.55 -0.6}{\sqrt{\frac{0.6(1-0.6)}{80}}}=-0.913$

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 assumed for this case is $\alpha=0.05$ . The next step would be calculate the p value for this test.

Since is a left tailed test the p value would be:

$p_v =P(z$

So the p value obtained was a very high value and using the significance level given $\alpha=0.05$ we see that $p_v>\alpha$ 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 proportion of graduate students show that only 44 students have ever done so is not significantly lower than 0.6

5 0

3 years ago