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Taya2010 [7]
3 years ago
5

Id really appreciate u if u answer this for em

Mathematics
1 answer:
Andreas93 [3]3 years ago
5 0
I think The answer is B
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Fill in the blank: In data analytics, a _____ refers to all possible data values in a certain dataset
earnstyle [38]

In data analytics, a p<u>opulation </u>refers to all possible data values in a certain dataset

Data analysis is a systematic computer-aided analysis of data or statistics. [1] It is used to discover, interpret, and convey meaningful patterns in the data. It also includes applying data patterns for effective decision-making.

It may be useful in areas where there is a lot of recorded information. The analysis relies on the simultaneous application of statistics, computer programming, and operations research to quantify performance.

Organizations can apply analytics to business data to describe, predict, and improve business performance. Areas within the analysis include descriptive analysis, diagnostic analysis, predictive analysis, prescriptive analysis, and cognitive analysis in particular. [2] Analysis can be applied in various fields such as marketing, management, finance, online systems, information security, and software services.

Learn more about the population here: brainly.com/question/25630111

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6 0
2 years ago
How can i solve it? I need to know how to solve iy
natita [175]
All of my resources are saying that this has no inverse.
6 0
3 years ago
Joanne was hired by a florist to make large boes for holidays wreaths each roll of ribbon has 20 yards joanne used 3 1/2 rolls o
alukav5142 [94]

Answer: Each bow needs 4.67 yards of ribbon.

Step-by-step explanation:

We know that each roll has 20 yd.

She used 3 + 1/2 rolls, then the total length used will be (3 + 1/2) times the length of each roll, this is:

(3 + 1/2)*20 yd = (6/2 + 1/2)*20yd = (7/2)*20yd = 70 yd.

With tose 70 yards of ribbon, she made 15 bows, then the length of ribbon used in each bow will be equal to the quotient between the total length used and the number of ribbons made with it, this is:

70yd/15 = 4.67 yards

Each bow needs 4.67 yards of ribbon.

6 0
3 years ago
The perimeter of a rectangle is 50 cm and the length is 4 cm longer than the width. Write and solve the system of equations usin
k0ka [10]
P=2(l+w)
50=2(4+w)
50=8+2w
collect the like terms 50-8=2w
42=2w divide both sides by 2
w=21

6 0
3 years ago
When engaging in weight-control (fitness/fat burning) types of exercise, a person is expected to attain about 60% of their maxim
Nimfa-mama [501]

Answer:

a) <em> standard error of the mean =10.06</em>

<em>b)  The margin of error  = 17.3982</em>

<em>c) 95% of confidence intervals are </em>

<em></em>(89.6018 ,124.3982)<em></em>

<em>d) Lower limit of 95% of confidence interval  = 89.6018</em>

<em>upper limit of 95% of confidence interval  = 124.3982</em>

<em>The Population mean is lies between in these intervals</em>

<u>Step-by-step explanation:</u>

<u><em>Step(i)</em></u><u>:-</u>

Given sample size 'n' = 20

Given sample mean was found to be 107 bpm with a standard deviation of 45 bpm.

<em>Sample mean             </em>x^{-} = 107 bpm<em></em>

<em>Sample standard deviation (S) = 45 bpm</em>

<em>a) standard error of the mean is determined by</em>

<em>     </em>S.E = \frac{S}{\sqrt{n} } = \frac{45}{\sqrt{20} }<em></em>

<em>     S.E = 10.06</em>

<em>b) The margin of error is determined by</em>

<em></em>M.E = \frac{t_{\alpha } X S }{\sqrt{n} }<em></em>

<em>The degrees of freedom  ν   </em>= n-1 =20-1=19<em></em>

<em>   </em>t_{\alpha } = 1.729  

<em></em>M.E = \frac{1.729X 45}{\sqrt{20} }<em></em>

<em></em>M.E = \frac{77.805 }{4.472 } = 17.3982<em></em>

<em>c) 95% of confidence intervals are determined by</em>

<em></em>(x^{-} - t_{\alpha } \frac{S}{\sqrt{n} } , x^{-} + t_{\alpha } \frac{S}{\sqrt{n} } )<em></em>

<em></em>(107 -  1.729\frac{45}{\sqrt{20} } , 107 + 1.729\frac{45}{\sqrt{20} } )<em></em>

<em></em>(107 -  17.3982 } , 107 +17.3982 )<em></em>

<em></em>(89.6018 ,124.3982)<em></em>

<em>d)  </em>

<em>Lower limit of 95% of confidence interval  = 89.6018</em>

<em>upper limit of 95% of confidence interval  = 124.3982</em>

<em>The Population mean is lies between in these intervals</em>

<em></em>

<em></em>

4 0
3 years ago
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