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Zina [86]
3 years ago
11

Work the following problem. Choose the correct answer. If an article costs a store $17.45 and if it sells the article for $22.95

, what percentage markup is it using?
%
Mathematics
2 answers:
Sedaia [141]3 years ago
5 0
(22.95−17.45)÷17.45)*100=31.52%
ruslelena [56]3 years ago
4 0

Answer:

The market price in percentage of the the article is 31.5%.

Step-by-step explanation:

Given : If an article costs a store $17.45 and if it sells the article for $22.95.

To find : What percentage markup is it using?

Solution :

The cost price of the article is CP=$17.45

The selling price of the article is SP=$22.95

Market price formula is

MP=\frac{SP-CP}{CP}

MP=\frac{22.95-17.45}{17.45}

MP=\frac{5.5}{17.45}

MP=0.315

The market price of the article is $0.315.

MP into percentage,

MP\%=MP\times 100\\MP\%=0.315\times 100\\MP\%=31.5\%

Therefore, The market price in percentage of the the article is 31.5%.

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Of 580580 samples of seafood purchased from various kinds of food stores in different regions of a country and genetically compa
Lubov Fominskaja [6]

Answer:

a) The 99% confidence interval would be given (0.204;0.296).

b) We have 99% of confidence that the true population proportion of all seafood sold in the country that is mislabeled or misidentified is between (0.204;0.296).  

c) No that's not true. Because the necessary assumptions and conditions for the confidence interval for the proportion are satisifed, so then we can use inferential statistics to interpret the interval to the population of interest.

Step-by-step explanation:

Part a

Data given and notation  

n=580 represent the random sample taken    

X represent the seafood sold in the country that is mislabeled or misidentified by the people

\hat p=0.25 estimated proportion of seafood sold in the country that is mislabeled or misidentified by the people

\alpha=0.01 represent the significance level (no given, but is assumed)    

p= population proportion of seafood sold in the country that is mislabeled or misidentified by the people

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The population proportion have the following distribution

p \sim N(p,\sqrt{\frac{\hat p(1-\hat p)}{n}})

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 99% confidence interval the value of \alpha=1-0.99=0.01 and \alpha/2=0.005, with that value we can find the quantile required for the interval in the normal standard distribution.

z_{\alpha/2}=2.58

And replacing into the confidence interval formula we got:

0.25 - 2.58 \sqrt{\frac{0.25(1-0.25)}{580}}=0.204

0.25 + 2.58 \sqrt{\frac{0.25(1-0.25)}{580}}=0.296

And the 99% confidence interval would be given (0.204;0.296).

Part b

We have 99% of confidence that the true population proportion of all seafood sold in the country that is mislabeled or misidentified is between (0.204;0.296).  

Part c

A government spokesperson claimed that the sample size was too​ small, relative to the billions of pieces of seafood sold each​ year, to generalize. Is this criticism​ valid?

No that's not true. Because the necessary assumptions and conditions for the confidence interval for the proportion are satisifed, so then we can use inferential statistics to interpret the interval to the population of interest.

4 0
3 years ago
Answer my math question I asked so many times I lost most of points by
kumpel [21]

Answer:

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Step-by-step explanation:

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I figured this out by looking at the point (6,0). To get 0 (y), you have to subtract 6. Knowing this, I subtracted 6 from the rest of the coordinates, leaving me with numbers that are able to be squared to get y. This led me to the equation x - 6 + x^2.

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Write the letter below for each fraction in the correct box?​
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The goal of a nutritional study was to compare the caloric intake of adolescents living in rural areas of the US with the calori
stiks02 [169]

Answer:

To give more clarity to the question, lets examine the attached back-to-back stem plot.

A)

Having examined the stem plot, we can using quick calculations, summarize that:

The mean (40.45 cal/kg) and median (41 cal/kg) daily caloric intake of ninth-grade students in the rural school is higher than the corresponding measures of center, mean (32.6 cal/kg) and median (32 cal/kg), for ninth-graders in the urban school.

The median and the mean for the students in the 9th grade in the urban school is lower than that of those of their contemporaries in the rural school. The respective medians and means are stated below:

Urban 9th Grade Students

Median = 32 cal/kg

Mean = 36 cal/kg

Rural 9th Grade Students

Median = 41 cal/kg

Mean = 41 cal/kg

Please note that all figures above have been approximated to the nearest whole number.

B)

It is unreasonable to generalize the findings of this study to all rural and urban 9th-grade students in the United States because the sample is too small compared to the target population size.

To allow for generalization, they would have to collect and analyze more samples say from every state within America.

Cheers!

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3 years ago
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