When two or more simple machines are combined they form

Which of these people is an engineer?

Debora [2.8K]

Answer:

a

Explanation:

the others dont make sense to be an engineer

8 0

3 years ago

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

3 years ago

What’s the most important benefit of maintaining a neutral posture

olga55 [171]

Neutral posture is essential for optimal wellbeing and functioning of the body. Holding the weight of the body The most important function of a neutral posture is to maintain the body in an upright position, supporting the body against gravity

7 0

3 years ago

A 50000 N plane has wings with a span of 30 m and a chord of 6 m. How much cargo can this plane carry while cruising at 550 km/h

soldi70 [24.7K]

Answer:

The amount of cargo the plane can carry is 8707.89 N

Explanation:

The surface area of the wings facing the air = 30×6×2 × sin(2.5) = 15.7 m²

The speed of the plane 550 km/h = 152.78 m/s

The volume of air cut through per second = 15.7 × 152.78 = 2399.07 m³

The mass of air = Volume × Density = 2399.07 × 0.37 = 887.65 kg

Weight of air = Mass × Acceleration due to gravity = 887.65 × 9.81 = 8707.89 N

Given that the plane is already airborne, the additional cargo the plane can carry is given by the available lift force of the plane.

The amount of cargo the plane can carry = 8707.89 N

8 0

4 years ago

A stream of liquid n-pentane flows at a rate of 50.4 L/min into a heating chamber, where it evaporates into a stream of air 15%

allochka39001 [22]

Answer:

(a) the fractional conversion of pentane achieved in the furnace is 90% conversion

(b) the volumetric flow rates (Umin) of the feed air is 256 x 10³ 1/m

(c) the volumetric flow rates (Lmin) of the gas leaving the condenser is 404.9 x 103 l/min

Explanation:

a) Molecular weight of pentane = 72.15 g/mol

density of liquid pentane = 626 kg/m3

Flow rate of feed liquid nitrogen = 50.4 l/min

= 626*50.4*10-3

= 31.55 kg/min

= 31.55/72.15 kmol/min

= 0.4372 kmol/min

Pentane existng the burner = 3.175 kg/min

Fractional conversion = (31.55 - 3.175)/ 31.55

= 0.9 = 90% conversion

b)

C₂H₅ + 8O₂ -----------> 5CO₂ + 6H₂O

From the Stoichiometric reaction,

8 mol of O2 are used for combustion of 1 mol of pentane

for 0.4372 kmol/min of pentane = 8 * 0.4372 kmol/min of Oxygen will be required

= 3.49 kmol/min of O2

amount of air will be = 3.49/0.21 = 16.62 kmol/min

15% excess air = 16.62*1.15 = 19.12 kmol/min

assuming air to be ideal gas

V = nRT / P.........(1)

V = 19.12 X 8.314 X 336 / 208.6

= 256 m³ = 256 x 10³ 1/m

c) Oxygen:

Amount of pentane consumed = (31.55 - 3.175) = 28.375 kg/min = 28.375/72.15 = 0.3932 kmol/min......(2)

Amount of O2 consumed = 8*0.3932 = 3.146 kmol/min

Amount of O2 fed by air = 0.21*19.12 = 4.0215 kmol/mim

unused O2 left = 4.0215 - 3.146

= 0.8755 kmol/min = 19.75*103 l/min............ (using (1))

Carbon Dioxide:

1 mol of pentane = 5 mol of CO2

0.3932 kmol/min of pentane = 5*0.3932 kmol/min of CO2..................(from (2)

= 1.966 kmol/min

= 44.362*103 l/min..........................(using (1))

Nitrogen:

v = 0.79 x 19.12 x 8.314 x 275 / 100.325

= 340.8 m³ / min = 340.8 x 10³ 1/min

Total volumetric flow rate of gases leaving the condenser = (340.8 + 44.36 + 19.75) x 103 l/min

= 404.9 x 103 l/min

6 0

3 years ago