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kenny6666 [7]
3 years ago
8

Please answer this correctly without making mistakes

Mathematics
1 answer:
Tomtit [17]3 years ago
6 0

Answer:

51\frac{4}{17}

Step-by-step explanation:

If we add all of the fractions together, we 'd get 55/17 of an hour. The question is to find how many hours she spent exercising. Well, for that we'd just need to see how many seventeens fit inside 55. We could divide, but that'd lead us to a really long, weird number.

Since 17*3=51, we know that in total, three seventeens fit inside 55. Yet, there's still remainders.

55-51=4

So, our answer would be 51 (how many 17s go into 55) and 4/17 (the remainder.)

Hope this helps!! <3 :)

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A modified box plot does not include the outliers in the whiskers, instead they are points outside of the whiskers

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Read 2 more answers
A recent study from the University of Virginia looked at the effectiveness of an online sleep therapy program in treating insomn
Ludmilka [50]

Answer:

1. The 99% confidence interval for the difference in average is -6.47377 < μ₁ - μ₂ < -11.34623

2. The possible issues in the calculations includes;

a. The confidence level used in the confidence interval can influence the result of the confidence interval observed

b. The sample size is small

Step-by-step explanation:

1. The number of adults with insomnia in the sample = 45

The number of adults that participated in the therapy, n₁ = 22

The number of candidates that served as control group, n₂ = 23

The average score for the for the 22 participants of the program, \overline x_1 = 6.59

The standard deviation for the 22 participants of the program, s₁ = 4.10

The average score for the for the 23 subjects in the control group, \overline x_2 = 15.50

The standard deviation for the 23 subjects in the control group, s₂ = 5.34

The confidence interval for unknown standard deviation, σ, is given by the following expression;

\left (\bar{x}_{1}- \bar{x}_{2}  \right )\pm t_{\alpha /2}\sqrt{\dfrac{s_{1}^{2}}{n_{1}}+\dfrac{s_{2}^{2}}{n_{2}}}

α = 1 - 0.99 = 0.01

α/2 = 0.005

The degrees of freedom, df = 22 - 1 = 21

t_{\alpha /2} = t_{0.005, \, 21} = 1.721

Therefore, we have;

\left (6.59- 15.5  \right )\pm1.721 \cdot \sqrt{\dfrac{4.10^{2}}{22}+\dfrac{5.34^{2}}{23}}

The 99% confidence interval for the difference in average is therefore given as follows;

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2. The possible issues in the calculations are;

a. The confidence level used in the confidence interval can influence the result of the confidence interval observed

b. The sample size is small

5 0
2 years ago
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