All categories

3 years ago

Show your work with long division please

Mathematics

1 answer:

Naya [18.7K]3 years ago

4 0

Answer:

Step-by-step explanation:

7/11 = no, add 0

70/11 = 6, so 0.6, remainder is 4, add 0

40/11 = 3, so 0.03, remainder is 7, add 0

70/11 = 6... and it goes on

<h3>The answer is 0.63 bar notation on both 6 and 3</h3>

You might be interested in

According to a Pew Research Center study, in May 2011, 35% of all American adults had a smart phone (one which the user can use

taurus [48]

Answer:

$z=\frac{0.4 -0.35}{\sqrt{\frac{0.35(1-0.35)}{300}}}=1.82$

$p_v =P(z>1.82)=0.034$

So the p value obtained was a very low value and using the significance level given $\alpha=0.1$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis

And the best conclusion would be:

There is enough evidence to show that more than 35% of community college students own a smart phone (Pvalue = 0.034).

And for the second case the correct system of hypothesis is:

H0: p = 0.078; Ha: p > 0.078

Step-by-step explanation:

Data given and notation

n=300 represent the random sample taken

$\hat p=\frac{120}{300}=0.4$ estimated proportion of college students that have a smart phone

$p_o=0.35$ is the value that we want to test

$\alpha=0.1$ represent the significance level

Confidence=90% or 0.90

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 proportion is >0.35.:

Null hypothesis: $p\leq 0.35$

Alternative hypothesis: $p > 0.35$

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:

$z=\frac{0.4 -0.35}{\sqrt{\frac{0.35(1-0.35)}{300}}}=1.82$

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 $\alpha=0.1$ . The next step would be calculate the p value for this test.

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

$p_v =P(z>1.82)=0.034$

So the p value obtained was a very low value and using the significance level given $\alpha=0.1$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis

And the best conclusion would be:

There is enough evidence to show that more than 35% of community college students own a smart phone (Pvalue = 0.034).

And for the second case the correct system of hypothesis is:

H0: p = 0.078; Ha: p > 0.078

4 0

2 years ago

Solve the following problem: –25 + 37 =

Lerok [7]

1. 12 is your answer

2. -33 is your answer

3 0

2 years ago

Read 2 more answers