Pipelines are a useful means of transporting oil because they offer low maintenance and dependable transportation for a narrow but important range of products.
<h3>What is a pipeline?</h3>
A pipeline is a system of connected pipelines that can be either underground or out in the environment. These pipelines are used to transport or distribute water, gas, and oil.
The options are attached
a. Pipelines provide jobs for consumers because of the resurgence of exploration and drilling in North America.
b. Pipelines are versatile, carrying more ton-miles than any other mode of transport over more than 2 million miles of pipeline.
c. Pipelines have more locations than water carriers.
d. Pipelines offer low maintenance and dependable transportation for a narrow but important range of products.
Thus, the correct option is d. Pipelines offer low maintenance and dependable transportation for a narrow but important range of products.
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Answer:
Correct option: B. 90%
Explanation:
The confidence interval is given by:
![CI = [\bar{x} - z\sigma_{\bar{x}} , \bar{x}+z\sigma_{\bar{x}} ]](https://tex.z-dn.net/?f=CI%20%3D%20%5B%5Cbar%7Bx%7D%20-%20z%5Csigma_%7B%5Cbar%7Bx%7D%7D%20%2C%20%5Cbar%7Bx%7D%2Bz%5Csigma_%7B%5Cbar%7Bx%7D%7D%20%5D)
If
is 190, we can find the value of
:



Now we need to find the value of
:


So the value of z is 1.71.
Looking at the z-table, the z value that gives a z-score of 1.71 is 0.0436
This value will occur in both sides of the normal curve, so the confidence level is:

The nearest CI in the options is 90%, so the correct option is B.
Answer:
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2. 
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4. 
Explanation:
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Answer:Topographic map. Contour line. Learning Objectives. After completing this chapter, you will be able to: □ Define civil engineering and civil drafting.
Explanation:
Explanation:
A.
H = Aeσ^4
Using the stefan Boltzmann law
When we differentiate
dH/dT = 4AeσT³
dH/dT = 4(0.15)(0.9)(5.67)(10^-8)(650)³
= 8.4085
Exact error = 8.4085x20
= 168.17
H(650) = 0.15(0.9)(5.67)(10^-8)(650)⁴
= 1366.376watts
B.
Verifying values
H(T+ΔT) = 0.15(0.9)(5.67)(10)^-8(670)⁴
= 1542.468
H(T+ΔT) = 0.15(0.9)(5.67)(10^-8)(630)⁴
= 1205.8104
Error = 1542.468-1205.8104/2
= 168.329
ΔT = 40
H(T+ΔT) = 0.15(0.9)(5.67)(10)^-8(690)⁴
= 1735.05
H(T-ΔT) = 0.15(0.9)(5.67)(10^-8)(610)⁴
= 1735.05-1059.83/2
= 675.22/2
= 337.61