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sweet [91]
3 years ago
10

If the nth term is 299find the value of n while arthmetic sequence is 13,24,35..

Mathematics
1 answer:
Art [367]3 years ago
3 0
Aritmetic
an=a1+d(n-1)
a1=first term
d=common differnce

common differnce is a term minus pervious term
24-13=11
d=11
a1=first term=13

13+11(n-1)
the nth term
299=13+11(n-1)
solve
distribute
299=13+11n-11
299=11n+2
minus 2 both sides
297=11n
divide both sides by 11
27=n
it is the 27th term
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<em>A. The F statistic increases as the differences among the sample means for the exercise groups increase.</em>

<em>C. The F statistic has 3 degrees of freedom in the numerator and 35 degrees of freedom in the denominator.</em>

<em>F. The F statistic is 9.2242.</em>

Step-by-step explanation:

Given that

Observations                     Samples                             Total

40-49             10       11              9         9                        39

The degrees of freedom is given by

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For denominator= n-k= 39-4= 35

F statistic is given by

F= MS between/ MS within

=11.632991 / 1.261143 = 9.2242

Where

Mean Square Between = MS between = 11.632991

Mean Square Within = MS within = 1.261143

<u>Hence the correct choices are  A, C and F</u>

<em>A. The F statistic increases as the differences among the sample means for the exercise groups increase.</em>

<em>C. The F statistic has 3 degrees of freedom in the numerator and 35 degrees of freedom in the denominator.</em>

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<em></em>

Suppose we have

Mean Square Between = MS between = 18.632991

Mean Square Within = MS within = 1.261143

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4 years ago
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3 years ago
Read 2 more answers
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lesya [120]

Answer:

0.1994 is the required probability.      

Step-by-step explanation:

We are given the following information in the question:

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Standard Deviation, σ = 5.3 pounds

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We are given that the distribution of weights is a bell shaped distribution that is a normal distribution.

Formula:

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Standard error due to sampling =

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0.1994 is the probability that the mean weight of a random sample of 20 boxers is more than 167 pounds

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