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mash [69]
3 years ago
7

A sanitary landfill has available space of 16.2 ha at an average depth of 10 m. Seven hundred sixty-five (765) cubic meters of s

olid waste are dumped at the site five days per week. This waste is compacted to twice its delivered density. Estimate the expected life of the landfill in years. (5 points)
Mathematics
1 answer:
AfilCa [17]3 years ago
6 0

Answer:

16.3 years

Step-by-step explanation:

1 ha = 10,000 m^2

The total capacity of the landfill is:

C = 16.2*10,000*10 = 1,620,000\ m^3

If waste is compacted to twice its delivered density, its volume is half of the delivered volume. Assuming that a year has 52 weeks, the volume of compacted solid waste dumped per year is:

V=52*(5*0.5*765)\\V=99,450\ m^3

The expected life of the landfill is given by its capacity divided by the yearly volume:

L=\frac{C}{V}=\frac{1,620,000}{99,450} \\L=16,3\ years

The landfill has an expected life of 16.3 years.

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7 0
3 years ago
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Jerry works for a cable television company. He must sell at least 10 television packages over the next week. Customers have two
LenKa [72]

Answer:

sefafaefawe

Step-by-step explanation:

6 0
3 years ago
2.5 x 1 1/10 x 9/3 / 1/2
Allisa [31]
4.125 is the answer
8 0
3 years 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

5 0
2 years ago
D=a/a+12*M. The adult weighs 75 kg. Calculate the adults weight in kilograms to pounds. Round final answer 2 decimal places
andrew-mc [135]

An adult with a weight of 75 kilograms have an <em>equivalent</em> weight of 165.60 pounds.

<h3>How to convert kilograms to pounds</h3>

Herein we have an application of <em>unit</em> conversions between <em>weight</em> units, from kilograms to pounds. Unit conversions follow this <em>direct proportional</em> formula:

y = k · x

Where:

x - Weight in kilograms

y - Weight in pounds

k - Conversion ratio

If we know that x = 75 kg and k = 2.208 lb/kg, then the weight of the adult in pounds is:

y = (2.208 lb/kg) · (75 kg)

y = 165.6 lb

An adult with a weight of 75 kilograms have an <em>equivalent</em> weight of 165.60 pounds.

To learn more on weight units: brainly.com/question/18762697

#SPJ1

5 0
1 year ago
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