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3 years ago

Factor -1/4 out of -1/2-5/4y

1 answer:

3 0

When we factor out something, we divide the factor by the equation. For example if we were to factor 3 out of 6, It will be 3(2). We got this answer by dividing 3 and 6. Using this concept:

$\frac{-1}{4}[( \frac{-1}{2} / \frac{-1}{4}) - ( \frac{5}{4} y / \frac{-1}{4})]$

Simplifying:

$\frac{-1}{4}(2 + 5y)$

Hope this helps!

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According to the University of Nevada Center for Logistics Management, of all merchandise sold in the United States gets returne

Step2247 [10]

Answer:

a) $\hat p=\frac{12}{80}=0.15$ estimated proportion of items that were returned

b) The 95% confidence interval would be given (0.0718;0.228).

c) Using a significance level assumed $\alpha=0.05$ we see that $p_v$ so we have enough evidence at this significance level to reject the null hypothesis. And on this case makes sense the claim that the proportion of returns at the Houston store significantly different from the returnsfor the nation as a whole.

Step-by-step explanation:

Assuming:

According to the University of Nevada Center for Logistics Management, 6% of all mer-chandise sold in the United States gets returned. Houston department store sampled 80 items sold in January and found that 12 of the items were returned.

Data given and notation

n=80 represent the random sample taken

X=12 represent the items that were returned

$\hat p=\frac{12}{80}=0.15$ estimated proportion of items that were returned

$\alpha=0.05$ represent the significance level (no given, but is assumed)

Confidence =0.95 or 95%

p= population proportion of items that were returned

a. Construct a point estimate of the proportion of items returned for the population ofsales transactions at the Houston store

$\hat p=\frac{12}{80}=0.15$ estimated proportion of items that were returned

b. Construct a 95% confidence interval for the porportion of returns at the Houston store

The confidence interval would be given by this formula

$\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}$

For the 95% confidence interval the value of $\alpha=1-0.95=0.05$ and $\alpha/2=0.025$ , with that value we can find the quantile required for the interval in the normal standard distribution.

$z_{\alpha/2}=1.96$

And replacing into the confidence interval formula we got:

$0.15 - 1.96 \sqrt{\frac{0.15(1-0.15)}{80}}=0.0718$

$0.15 + 1.96 \sqrt{\frac{0.15(1-0.15)}{80}}=0.228$

And the 95% confidence interval would be given (0.0718;0.228).

c. Is the proportion of returns at the Houston store significantly different from the returns for the nation as a whole? Provide statistical support for your answer.

We need to conduct a hypothesis in order to test the claim that the population proportion differs significantly to the USA proportion of 6% or no. We have the following system of hypothesis :

Null Hypothesis: $p = 0.06$

Alternative Hypothesis: $p \neq 0.06$

We assume that the proportion follows a normal distribution.

This is a two tailed test for the proportion .

The One-Sample Proportion Test is "used to assess whether a population proportion $\hat p$ is significantly (different,higher or less) from a hypothesized value $p_o$ ".

Check for the assumptions that he sample must satisfy in order to apply the test

a)The random sample needs to be representative: On this case the problem no mention about it but we can assume it.

b) The sample needs to be large enough

$np_o =80*0.15=12>10$

$n(1-p_o)=80*(1-0.15)=68>10$

Calculate the statistic

The statistic is calculated with the following formula:

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

On this case the value of $p_o=0.06$ is the value that we are testing and n = 80.

$z=\frac{0.15 -0.06}{\sqrt{\frac{0.06(1-0.06)}{80}}}=3.390$

The p value for the test would be:

$p_v =2*P(z>3.390)=0.00070$

Using a significance level assumed $\alpha=0.05$ we see that $p_v$ so we have enough evidence at this significance level to reject the null hypothesis. And on this case makes sense the claim that the proportion of returns at the Houston store significantly different from the returnsfor the nation as a whole.

5 0

4 years ago

What is the value of the 4 in the number 17.884 give your answer as a fraction

aleksandrvk [35]

To find : What is the value of 4 in the number 17.884 as a fraction ? Therefore, the value of 4 in the number 17.884 as a fraction is four-thousandths .

6 0

3 years ago

Name ____________________________________________Date_______ Class __ Score_______

Svetradugi [14.3K]

Answer:

its very long but ill help you with the first answer:18.5753424657534247

Step-by-step explanation:

7 0

3 years ago

Using the fixed-time period inventory model, and given an average daily demand of 200 units, 4 days between inventory reviews, 5

KATRIN_1 [288]

Answer:

Correct option: B. About 1,686.

Step-by-step explanation:

The formula to compute the order quantity (Q) is:

$Q=(q_{d}\times (I+L))+(z\times\sigma_{I+L})-I_{n}$

Here

$q_{d}=average\ daily\ semand=200\\I = Inventory\ review\ time=4\\L=lead\ time=5\\\sigma_{I+L}=standard\ deviation\ over\ the\ review\ and\ lead\ time=3\\I_{n}=number\ of\ units\ of\ inventory\ on\ hand=120$

Compute the order quantity as follows:

$Q=(q_{d}\times (I+L))+(z\times\sigma_{I+L})-I_{n}\\=(200\times(4+5))+(1.96\times 3)-120\\=1800+5.88-120\\=1685.88\\\approx1686$

Thus, the order quantity was about 1,686.

5 0

3 years ago

How many square feet of outdoor carpet will we need for this hole?

Alex73 [517]

Answer:

Idk if it’s asking for the carpet for the hole but if it’s asking how much carpet to fill the hole the answer is 4, but if it’s asking how much carpet to cover the green area the answer is 44 feet

Step-by-step explanation:

6 0

3 years ago