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Levart [38]
2 years ago
13

The Environmental Protection Agency (EPA) has standards and regulations that says that the lead level in soil cannot exceed the

limit of 400 parts per million (ppm) in public play areas designed for children. Mike is an inspector, and he takes 50 randomly selected soil samples from a site where they are considering building a playground.
These data show a sample mean of x = 390.25 ppm and standard deviation of Sx = 30.5 ppm. Answer the following:

1. Determine the critical t value and along with the confidence interval for 95% level of

confidence. 2. What does this interval suggest?
Engineering
1 answer:
DENIUS [597]2 years ago
3 0

Answer:

See below

Explanation:

<u>Check One-Sample T-Interval Conditions</u>

Random Sample? √

Sample Size ≥30? √

Independent? √

Population Standard Deviation Unknown? √

<u>One-Sample T-Interval Information</u>

  • Formula --> CI=\bar{x}\pm t^*(\frac{S_x}{\sqrt{n}})
  • Sample Mean --> \bar{x}=390.25
  • Critical Value --> t^*=2.0096 (given df=n-1=50-1=49 degrees of freedom at a 95% confidence level)
  • Sample Size --> n=50
  • Sample Standard Deviation --> S_x=30.5

<u>Problem 1</u>

The critical t-value, as mentioned previously, would be t^*=2.0096, making the 95% confidence interval equal to CI=\bar{x}\pm t^*(\frac{S_x}{\sqrt{n}})=390.25\pm2.0096(\frac{30.5}{\sqrt{50}})\approx\{381.5819,398.9181\}

This interval suggests that we are 95% confident that the true mean levels of lead in soil are between 381.5819 and 398.9181 parts per million (ppm), which satisfies the EPA's regulated maximum of 400 ppm.

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Air is to be heated steadily by an 8-kW electric resistance heater as it flows through an insulated duct. If the air enters at 5
Furkat [3]

To solve this problem it is necessary to apply the concepts related to the heat exchange of a body.

By definition heat exchange in terms of mass flow can be expressed as

W = \dot{m}c_p \Delta T

Where

C_p = Specific heat

\dot{m}= Mass flow rate

\Delta T = Change in Temperature

Our values are given as

C_p = 1.005kJ/kgK \rightarrow Specific heat of air

T_1 = 50\°C

\dot{m} = 2kg/s

W = 8kW

From our equation we have that

W = \dot{m}c_p \Delta T

W = \dot{m}c_p (T_2-T_1)

Rearrange to find T_2

T_2 = \frac{W}{\dot{m}c_p}+T_1

Replacing

T_2 = \frac{8}{2*1.005}+(50+273)

T_2 = 326.98K \approx 53.98\°C

Therefore the exit temperature of air is 53.98°C

6 0
3 years ago
5. Identify the pros and cons of<br> manufactured siding.
lapo4ka [179]
Aluminum
Aluminum siding is out of date and really has no advantages in today’s market. It dents, it’s difficult to keep clean, and it’s hard to update it once it’s installed. Aluminum siding is not recommended, and most professional contractors won’t even install it.

Vinyl
Today’s mainstream siding material is vinyl. It is the most economical type of siding, coming in at $4 to $5 per square foot.

Available in a wide variety of colors, vinyl siding not only protects your home, it can make a statement. You can choose from whites, ivories, grays, browns, reds, greens and even blues — there’s a color for everyone.

In addition, vinyl is a complete self-flushing water-exclusion system. It is made to be water repellent and durable in any kind of weather. Its lightweight footprint makes it easy to install, which minimizes expenses.

Vinyl is low-maintenance and requires only periodic cleaning with a power washer every few years – more if you want it to really shine.

Premium vinyl
A step up from regular vinyl, premium vinyl gives you a more wood-like appearance. It’s usually a thicker vinyl that looks more rigid because it has no oil canning, as lighter vinyl does.

The upgrade to premium gives you more strength and longer lifespan. Many premium vinyl sidings also have integrated insulation or r-values (insulation grade) of 5 to 6.

Premium vinyl cost is generally higher than vinyl, at around $7 to 8 per square foot. Premium vinyl also comes in a rainbow of colors for customization and personalization

Natural wood
Wood siding provides a more rustic look for your home, with a more customized result.

The siding can be in long horizontal pieces or it can be shaped to suit your individual design needs. That’s the beauty of wood. You may want the look of clapboard – overlapping pieces of wood installed horizontally – or shingles – smaller pieces of wood overlapping like a shingled roof. Wood siding is a good choice for this façade.

Fiber cement siding
Finally, you may hear about fiber cement siding. Today’s version is a combination of wood pulp, cement, clay and sand. It can be made to look like wood siding, and it is installed and wears much like wood siding.

This is different from the fiber cement siding that was used in homes built prior to the late 1980s. Those older sidings contain asbestos, and any work with them should be done by a specialized asbestos-removal professional. Today’s fiber cement siding is safer and longer-lasting, if you’re willing to pay the price.

Fiber cement siding can be ordered pre-colored, eliminating the need to paint the siding. If you prefer to paint it, that is still an option, because the material does accept paint rather easily. Fiber cement siding requires a periodic recaulk of butt joints at trim terminations, which is a little more maintenance than a vinyl siding.

When to replace siding
When should you replace your siding? It should be replaced if it has any kind of water damage, mold or rotting. If it hasn’t been properly maintained, it might be time for new siding. And, of course, if you simply want a new look, new siding can be a side show for the neighborhood.
5 0
3 years ago
The normal stress at gage H calculated in Part 1 includes four components: an axial component due to load P, σaxial, P, a bendin
Degger [83]

Answer:

hello your question has some missing information attached to the answer is the missing component

Answer : αaxial,p = -6.034 ksi ( compressive )

             αbend,p = 19.648 ksi ( tensile )

Explanation:

αaxial, p = \frac{-p}{A}   equation 1

αbend, p = \frac{(P*A)*\frac{d}{2} }{I_{z} } equation 2

P = load = 35 kips

A = area of column = 5.8 in^{2}

d = column cross section depth = 9.5 in

I_{Z} = 55.0 in^{4}

Hence equation 1 becomes

αaxial,p = -35 / 5.8 = - 6.034 ksi ( compressive )

equation 2 becomes

αbend, p = \frac{(35*6.5)(\frac{9.2}{2}) }{55} = + 19.648 ksi ( tensile )

7 0
3 years ago
The engine of a 2000kg car has a power rating of 75kW. How long would it take (seconds) to accelerate from rest to 100 km/hr at
Delvig [45]

Answer: 10.29 sec.

Explanation:

Neglecting drag and friction, and at road level , the energy developed during the time the car is accelerating, is equal to the change in kinetic energy.

If the car starts from rest, this means the following:

ΔK = 1/2 m*vf ²

As Power (by definition) is equal to Energy/Time= 75000 W= 75000 N.m/seg, in order to get time in seconds, we need to convert 100 km/h to m/sec first:

100 (Km/h)*( 1000m /1 Km)*(3600 sec/1 h)= 27,78 m/sec

Now, we calculate the change in energy:

ΔK= 1/2*2000 Kg. (27,78)² m²/sec²= 771,728 J

<h2>If P= ΔK/Δt, </h2><h2>Δt= ΔK/P= 771,728 J / 75,000 J/sec= 10.29 sec.</h2>
4 0
3 years ago
Name the famous engineer in the world​
fgiga [73]

Answer:

Nikola Tesla

Explanation:

5 0
3 years ago
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