Question

n analysis is performed of a company’s daily revenues, which are$2.54million,$0.87 million

Analyzing functions
ANSWERED
asked 2021-02-21
n analysis is performed of a company’s daily revenues, which are $2.54million, $0.87 million, $$$1.66 million, and so on.

A. A boxplot could not be used to represent this data set. A pie chart could not be used to represent this data set.

B. A boxplot could be used to represent this data set. A pie chart could not be used to represent this data set.

C. A boxplot could not be used to represent this data set. A pie chart could be used to represent this data set.

D. A boxplot could be used to represent this data set. A pie chart could be used to represent this data set.

Expert Answers (1)

2021-02-22
For statement: An analysis is performed of a companys daily revenues, which are $2.54 million, $0.87 million, $1.66 million, and so on, the answer is B. Now we will explain why. We use box plot to analyse this data becatise there we can represent sample quartiles, upper and lower, sample median and minimum and maximum of given data. A pie chart is a circular statistical graphic, which is divided into slices to illustrate numerical proportion. In a pie chart, the arc length of each slice is: proportional to the quantity it represents. The answer is B.
46
 
Best answer

expert advice

Have a similar question?
We can deal with it in 3 hours

Relevant Questions

asked 2020-12-17

Error Analysis A student evaluated \(- 4 + x\ for\ x = - 9\) and got an answer of 5. What might the student have done wrong?

asked 2021-03-09

Error Analysis A student solved the equation \(2\cos\theta=\sqrt{2}\ for\ 0\leq\theta<2\pi\) and got \(\theta=(\pi/4)\). What was her error?

asked 2020-11-23

Geographical Analysis (Oct. 2006) published a study of a new method for analyzing remote-sensing data from satellite pixels in order to identify urban land cover. The method uses a numerical measure of the distribution of gaps, or the sizes of holes, in the pixel, called lacunarity. Summary statistics for the lacunarity measurements in a sample of 100 grassland pixels are \(\bar{x}=225\ and\ s=20s=20\). It is known that the mean lacunarity measurement for all grassland pixels is 220. The method will be effective in identifying land cover if the standard deviation of the measurements is 10% (or less) of the true mean (i.e., if the standard deviation is less than 22).

a. Give the null and alternative hypotheses for a test to determine whether, in fact, the standard deviation of all grassland pixels is less than 22.

b. A MINITAB analysis of the data is provided below. Locate and interpret the p-value of the test. Use \(\alpha=0.10\). Test for One Standard Deviation Method Null hypothesis \(\Sigma = 22\) Method Alternative hypothesis \(\Sigma = < 22\) The standard method is only for the normal distribution. Statistics NStDevVariance 10020.0400 Tests

asked 2021-06-09

Error Analysis On your homework, you write that the missing term in the arithmetic sequence 31, _, 41, ... is \(\displaystyle{35}½\). Your friend says the missing term is 36. Who is correct? What mistake was made?

...