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FrozenT [24]
3 years ago
13

What’s the answer I don’t know how to do it

Mathematics
1 answer:
weeeeeb [17]3 years ago
8 0
The answer would be 5. If you look at the table, look at what they are multiplying by. If you don't know how to do that, all you do is divide the bigger number by the smaller one and then you will get your answer

Hope this helps :)
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Please Help , Thank You
olya-2409 [2.1K]
I'll go with A and B 

Hope this helps 
6 0
4 years ago
Read 2 more answers
WILL MARK BRAINLIEST!!!!!!!!
SVETLANKA909090 [29]

Data:

15 16 14 15 19 17

n=6 points

sum is 96

mean is 96/6 = 16

Now we look at the absolute deviations, each of which is the absolute value of a score minus the mean, basically the distance of the score to the mean .

Scores   15 16 14 15 19 17

AbsDev    1  0  2  1    3  1

The sum of the absolute deviations is 8 and there are six of them so the

Mean Absolute Deviation = 8/6 = 4/3

Answer: 2.    8/6

7 0
4 years ago
What makes the sentence true 1\2=?/12
bagirrra123 [75]
It would be 1/2 = 6/12
6 0
4 years ago
Which of the following is equivalent to log 7^ 50 rounded to three decimal places?
otez555 [7]
7^50
= 1.798 × 10^42

i am a mathematics teacher. if anything to ask please pm me
4 0
3 years ago
1. How many variables are involved in the chi-square test?
GarryVolchara [31]

The result of the respective questions are:

  • This chi-square test only takes into consideration one variable.
  • The  type of chi-square test this is is a Goodness of Fit
  • df= 3
  • NO

<h3>How many variables are involved in the chi-square test?</h3>

a)

This chi-square test only takes into consideration one variable.

b)

The  type of chi-square test this is, is a Goodness of Fit

To test the hypothesis, we must determine whether the actual data conform to the assumed distribution.

The "Goodness-of-Fit" test is a statistical hypothesis test that determines how well the data that was seen resembles the data that was predicted.

c)

Parameter

n = 4

Therefore

Degrees of freedom

df= n - 1

df= 4 - 1

df= 3

d)

In conclusion

Parameters

\alpha = 0.05

df = 3

Hence

Critical value = 7.814728

Test statistic = 6.6

Test statistic < Critical value, .

NO, the result of this test is not statistically significant.

Read more about Probability

brainly.com/question/11234923

#SPJ1

3 0
2 years ago
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