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Ksju [112]
3 years ago
9

A six-lane freeway (three lanes in each direction) in rolling terrain has 10-ft lanes and obstructions 5 ft from the right edge.

There are zero ramps within three miles upstream of the segment midpoint and one ramp within three miles downstream of the segment midpoint. The traffic stream has a peak hour factor of 0.84, peak-hour volume of 2500 vehicles, and 4% recreational vehicles. What is the level of service
Engineering
1 answer:
Anon25 [30]3 years ago
3 0
That is too hard but u got that cuz i believe in you!!!
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A random sample of 5 hinges is selected from a steady stream of product from a punch press, and the a. b. proportion nonconformi
professor190 [17]
<h2>Answer with explanation:</h2>

Given : Sample size : n= 5

The proportion nonconforming : p= 0.10

Binomial probability formula :-

P(x)=^nC_x p^{x}(1-p)^{n-x}

The probability of zero nonconforming unit in the sample :-

P(0)=^5C_0 (0.10)^{0}(1-0.1)^{5}\\\\=(1)(0.9)^5\ \ [ \because\ ^nC_0=1]=0.59049

∴ The probability of zero nonconforming unit in the sample= 0.59049

The probability of one nonconforming unit in the sample :-

P(1)=^5C_1 (0.10)^{1}(0.9)^{4}\\\\=(5)(0.1)(0.9)^5\ \ [ \because\ ^nC_1=n]=0.295245

∴ The probability of one nonconforming unit in the sample=0.295245

The  probability of 2 or more nonconforming units in the sample :-

P(X\geq2)=1-(P(0)+P(1))=1-(0.59049+0.295245)\\\\=1-0.885735=0.114265

∴ The  probability of 2 or more nonconforming units in the sample=0.114265

6 0
3 years ago
A manager has a list of items that have been sorted according to an item ID. Some of them are duplicates. She wants to add a cod
ruslelena [56]

Answer:

The solution code is written in Python:

  1. items = [{"id": 37697, "code": ""},{"id": 37698, "code": ""},{"id": 37699, "code": ""},{"id": 37699, "code": ""}, {"id": 37699, "code": ""},
  2. {"id": 37699, "code": ""},{"id": 37699, "code": ""},{"id": 37699, "code": ""},{"id": 37700, "code": ""} ]
  3. items[0]["code"] = 1
  4. for i in range(1, len(items)):
  5.    if(items[i]["id"] == items[i-1]["id"]):
  6.        items[i]["code"] = items[i-1]["code"] + 1
  7.    else:
  8.        items[i]["code"] = 1
  9. print(items)

Explanation:

Firstly, let's create a list of dictionary objects. Each object holds an id and a code (Line 1-2). Please note all the code is initialized with zero at the first beginning.

Next, we can assign 1 to the <em>code</em> property of items[0] (Line 4).

Next, we traverse through the items list started from the second element (Line 6). We set an if condition to check if the current item's id is equal to the previous item (Line 7). If so, we assign the previous item's code + 1 to the current item's code (Line 8). If not, we assign 1 to the current item's code (Line 10).

At last, we print out the item (Line 12) and we shall get

[{'id': 37697, 'code': 1}, {'id': 37698, 'code': 1}, {'id': 37699, 'code': 1}, {'id': 37699, 'code': 2}, {'id': 37699, 'code': 3}, {'id': 37699, 'code': 4}, {'id': 37699, 'code': 5}, {'id': 37699, 'code': 6}, {'id': 37700, 'code': 1}]

 

6 0
3 years ago
9) A construction company employs 2 sales engineers. Engineer 1 does the work in estimating cost for 70% of jobs bid by the comp
Irina18 [472]

The probability that the error occurred when Engineer 2 made the mistake is 0.462 on the other hand the probability that the error occurred when the engineer 1 made the mistake is 0.538

Explanation:

Let E_{1} denote the event that the 1st  engineer  does the work.so we write

P(E)_{1}=0.7

Let E_{2} denote the event that the 2nd engineer  does the work .So we write

P(E)_{2}=0.3

Let O denote the event during which the error occurred .so we write

P(O/E_{1} )=0.02(GIVEN)

P(O/E_{2} )=0.04(GIVEN)

  • The probability that the error occurred when the first engineer performed the work is P(E_{1} /O)
  • The probability that the error occurred when the first engineer performed the work is P(E_{2} /O)

Now we need to find when did the error in the work occur so we will compare the probability of the work done by <u>engineer 1 </u>and <u>engineer 2 </u>

<u></u>

<u>lets find the Probability of the Engineer 1</u>

<u>Using Bayes theorem,we get</u>

<u></u>

<u></u>P(E_{1} /O) =0.02*0.7/0.02*0.7+0.04*0.3 = 0.014/0.026=0.538

<u>lets find the Probability of the Engineer 2</u>

<u></u>P(E_{2} /O) =0.04*0.3/0.02*0.7+0.04*0.3=0.012/0.026=0.462

Since ,0.462<0.538 so it is more prominent that the Engineer 1 did the work when the error occurred

3 0
3 years ago
Identify SIX (6) objectives of maintenance.<br>​
Rasek [7]

Answer:

to optimize the reliability of equipment and infrastructure;

- to ensure that equipment and infrastructure are always in good condition;

- to carry out prompt emergency repair of equipment and infrastructure so as to secure the best possible availability for production;

- to enhance, through modifications, extensions, or new low-cost items, the productivity of existing equipment or production capacity;

- to ensure the operation of equipment for production and for the distribution of energy and fluids;

- to improve operational safety;

- to train personnel in specific maintenance skills;

- to advise on the acquisition, installation and operation of machinery;

- to contribute to finished product quality;

- to ensure environmental protection.

Explanation:

pick whichever you want

8 0
2 years ago
What is the color of an original apple
marta [7]

Answer:

red

Explanation:

ok boomer

5 0
3 years ago
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