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mina [271]
2 years ago
9

Which phase of inferential statistics is sometimes considered to be the most crucial because errors in this phase are the most d

ifficult to correct?
Mathematics
1 answer:
Alex_Xolod [135]2 years ago
7 0

The phase of inferential statistics which is sometimes considered to be the most crucial because errors in this phase are the most difficult to correct is "data gathering".

<h3>What is inferential statistics?</h3>

Inferential statistics are frequently employed to compare treatment group differences.

Some characteristics of inferential statistics are-

  • Inferential statistics compare treatments groups and make conclusions about the greater population of participants using measures from the experiment's sample of subjects.
  • Inferential statistics aids in the development of explanations for a condition or phenomenon.
  • It enables you to draw conclusions on extrapolations, which distinguishes it from descriptive statistics, which simply summarize the information that has been measured.
  • There are numerous varieties of inferential statistics, each with its own set of research design & sample characteristics.
  • To select the correct statistical test of their experiment, researchers should reference the numerous texts about experimental design and statistics.

To know more about the inferential statistics, here

brainly.com/question/4774586

#SPJ4

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There are eight different jobs in a printer queue. Each job has a distinct tag which is a string of three upper case letters. Th
Vikentia [17]

Answer:

a. 40320 ways

b. 10080 ways

c. 25200 ways

d. 10080 ways

e. 10080 ways

Step-by-step explanation:

There are 8 different jobs in a printer queue.

a. They can be arranged in the queue in 8! ways.

No. of ways to arrange the 8 jobs = 8!

                                                        = 8*7*6*5*4*3*2*1

No. of ways to arrange the 8 jobs = 40320 ways

b. USU comes immediately before CDP. This means that these two jobs must be one after the other. They can be arranged in 2! ways. Consider both of them as one unit. The remaining 6 together with both these jobs can be arranged in 7! ways. So,

No. of ways to arrange the 8 jobs if USU comes immediately before CDP

= 2! * 7!

= 2*1 * 7*6*5*4*3*2*1

= 10080 ways

c. First consider a gap of 1 space between the two jobs USU and CDP. One case can be that USU comes at the first place and CDP at the third place. The remaining 6 jobs can be arranged in 6! ways. Another case can be when USU comes at the second place and CDP at the fourth. This will go on until CDP is at the last place. So, we will have 5 such cases.

The no. of ways USU and CDP can be arranged with a gap of one space is:

6! * 6 = 4320

Then, with a gap of two spaces, USU can come at the first place and CDP at the fourth.  This will go on until CDP is at the last place and USU at the sixth. So there will be 5 cases. No. of ways the rest of the jobs can be arranged is 6! and the total no. of ways in which USU and CDP can be arranged with a space of two is: 5 * 6! = 3600

Then, with a gap of three spaces, USU will come at the first place and CDP at the fifth. We will have four such cases until CDP comes last. So, total no of ways to arrange the jobs with USU and CDP three spaces apart = 4 * 6!

Then, with a gap of four spaces, USU will come at the first place and CDP at the sixth. We will have three such cases until CDP comes last. So, total no of ways to arrange the jobs with USU and CDP three spaces apart = 3 * 6!

Then, with a gap of five spaces, USU will come at the first place and CDP at the seventh. We will have two such cases until CDP comes last. So, total no of ways to arrange the jobs with USU and CDP three spaces apart = 2 * 6!

Finally, with a gap of 6 spaces, USU at first place and CDP at the last, we can arrange the rest of the jobs in 6! ways.

So, total no. of different ways to arrange the jobs such that USU comes before CDP = 10080 + 6*6! + 5*6! + 4*6! + 3*6! + 2*6! + 1*6!

                    = 10080 + 4320 + 3600 + 2880 + 2160 + 1440 + 720

                    = 25200 ways

d. If QKJ comes last then, the remaining 7 jobs can be arranged in 7! ways. Similarly, if LPW comes last, the remaining 7 jobs can be arranged in 7! ways. so, total no. of different ways in which the eight jobs can be arranged is 7! + 7! = 10080 ways

e. If QKJ comes last then, the remaining 7 jobs can be arranged in 7! ways in the queue. Similarly, if QKJ comes second-to-last then also the jobs can be arranged in the queue in 7! ways. So, total no. of ways to arrange the jobs in the queue is 7! + 7! = 10080 ways

5 0
4 years ago
Could someone plz help and show work? Thanks
arsen [322]

Answer:

3 cm

Step-by-step explanation:

8 0
3 years ago
Read 2 more answers
Which of these cannot be an 2-value?<br> O A. 0.75<br> O B. -0.75<br> O c. 1<br> O D.O
Kobotan [32]
The answer is B. -0.75
4 0
4 years ago
Help resolve this please
Leokris [45]

Answer:

this is late but 1176?

Step-by-step explanation:

5 0
3 years ago
Read 2 more answers
Please help! A rectangle with an area of 104 square centimeters has a length and width in a ratio of 13:2.
Ghella [55]

Answer:

length = 26 cm

width = 4 cm

Step-by-step explanation:

A = 104 cm²

A = length x width

ratio of length and width = 13 : 2

total ratio of length and width is 15 = a

so ratio of length is (13/15)a and width (2/15)a

A = length x width = (13/15)a x (2/15)a = (26/225)a² = 104

a² = 104 x 225 : 26 = 900

a = 30cm

length = (13/15)a = 13/15 x 30 = 26 cm

width = (2/15)a = 2/15 x 30 = 4 cm

4 0
3 years ago
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