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2 years ago

Find each sum.Simplify if possible

Mathematics

2 answers:

Alex Ar [27]2 years ago

8 0

Number 2 is 7 9/8 but its an improper fraction so you'll have to reduce that since im having trouble doing that.

Number 5 is 9 11/12

Hope this helps!

Dmitrij [34]2 years ago

6 0

2) 6 3/4 is = to 6 6/8 1 3/8 + 6 6/8= 8 1/8 The answer is 8 1/3 5) 3 1/4 ×3=3 3/12 6 2/3 ×4=6 8/12 3 3/12+ 6 8/12= 9 11/12 The answer is 9 11/12

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x= -6 & x= 2.5

Step-by-step explanation:

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A 56-foot piece of siding is cut into three pieces so that the second piece is twice as long as the first piece and the third pi

ELEN [110]

The lengths of the first, second and third piece are respectively 8 ft, 16 ft and 32 ft

<h3>How to solve Algebra word Problems?</h3>

Let the length of the first piece be x.

Let the length of the second piece be 2x

Let the length of the thrid piece be 4x

Since the total length of the piece is 56 ft, then we can say that;

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1 year ago

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;

-6.47377 < μ₁ - μ₂ < -11.34623

Therefore, there is considerable evidence that the participants in the survey had lower average score than the subjects in the control group

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

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2 years ago