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Varvara68 [4.7K]
3 years ago
5

How are industries related to systems

Engineering
1 answer:
REY [17]3 years ago
7 0

Answer:

The industrial systems of the future are seen as complex systems, composed of vast numbers of devices, interacting with each other and with enterprise systems continuously.

summary:

they are related because they ARE a system. well, a type.

hope this helps!! :)

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One unethical decision won't cause any harm.( true) or (false)
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Thermoplastics burn upon heating. a)-True b)- false?
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Suppose a consumer advocacy group would like to conduct a survey to find the proportion p of consumers who bought the newest gen
Anvisha [2.4K]

Answer:

a) Sample size = 1691

b) 95% Confidence Interval = (0.3696, 0.4304)

Explanation:

(a) How large a sample n should they take to estimate p with 2% margin of error and 90% confidence?

The margin of error is given by

MoE = z \cdot \frac{\sqrt{p(1-p)} }{\sqrt{n} }  \\\\

Where z is the corresponding z-score for 90% confidence level

z = 1.645 (from z-table)

for p = 0.50 and 2% margin of error, the required sample size would be

n = \frac{1.645^{2} \cdot 0.50(1-0.50)}{0.02^{2}}  \\\\n = \frac{0.6765}{0.0004}  \\\\n = 1691\\

(b) The advocacy group took a random sample of 1000 consumers who recently purchased this mobile phone and found that 400 were happy with their purchase. Find  a 95% confidence interval for p.

The sample proportion is

p = 400/1000

p = 0.40

z = 1.96 (from z-table)

n = 1000

The confidence interval is given by

CI = p \pm z \cdot \sqrt{\frac{p(1-p)}{n} } \\\\CI = 0.40 \pm 1.96 \cdot \sqrt{\frac{0.40(1-0.40)}{1000} } \\\\CI = 0.40 \pm 1.96 \cdot 0.01549 \\\\CI = 0.40 \pm 0.0304 \\\\CI = 0.40 - 0.0304 \: and \: 0.40 + 0.0304\\\\CI = (0.3696 ,\:  0.4304)

Therefore, we are 95% confident that the proportion of consumers who bought the newest generation of mobile phone were happy with their purchase is within the range of (0.3696, 0.4304)

What is Confidence Interval?

The confidence interval represents an interval that we can guarantee that the target variable will be within this interval for a given confidence level.  

3 0
4 years ago
A vehicle of 1 200 kg is moving at a speed of 40 km/h on an incline of 1 in 50. The total constant rolling and wind resistance i
Readme [11.4K]

Answer:11.602 KW

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mass of vehicle\left ( m\right )=1200 kg

speed=40Km/h

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\eta =80%

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Net force to overcome by engine is

F=Resistance + sin component of weight

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Where tan\theta =[tex]\frac{1}{50}

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power=F.v=835.38\frac{100}{9}

Engine Power=\frac{835.\frac{100}{9}}{\eta }=11.602 KW

4 0
4 years ago
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