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1 year ago

A carpenter needs 9.3 feet of rope for a project. How much will the carpenter spend if the rope costs $3.82 per foot?

Mathematics

1 answer:

Vlada [557]1 year ago

6 0

The carpenter spend 35.53 $ of rope for a project.

To find the cost of rope for a project.

What is arithmetic?

The branch of mathematics dealing with the properties and manipulation of numbers. a science that deals with the addition, subtraction, multiplication, and division of numbers. Fractions, decimals, percentages, fractions, square root, exponents, and other arithmetic operations are used to achieve mathematical simplifications.

Given that:

For a project carpenter needs 9.3 feet of rope.

Per foot of cost of the rope is $3.82.

So by calculation

1 foot of rope= $3.82.

9 foot of rope=9*3.82 $

9 foot of rope=35.53 $

So, the carpenter spend 35.53 $ of rope for a project.

Learn more about arithmetic here:

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Option is missing

O $35.53

O$44.84

O$45.84

O $355.30

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A recent study from the University of Virginia looked at the effectiveness of an online sleep therapy program in treating insomn

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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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