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S_A_V [24]
4 years ago
7

Searches related to Probability questions - A person frequents one of the two restaurants KARIM or NAZEER, choosing Chicken's it

em 70% of the time and fish's item 30% of the time. Regardless of where he goes , he orders Afghani Chicken 60% of his visits. (a) The next time he goes into a restaurants, what is the probability that he goes to KARIM and orders Afghani Chicken. (b) Are the two events in part a independent? Explain. (c) If he goes into a restaurants and orders Afghani Chicken, what is the probability that he is at NAZEER. (d) What is the probability that he goes to KARIM or orders Afghani Chicken or both?
Engineering
1 answer:
MakcuM [25]4 years ago
7 0

Answer:

a) 0.42

b) Independent

c) 30%

d) 0.88

Explanation:

Person chooses Chicken's item : 70% = 0.7

Person chooses fish's item : 30% = 0.3

Visits in which he orders Afghani Chicken = 60% = 0.6

a) Probability that he goes to KARIM and orders Afghani Chicken:

P = 0.7 * 0.6 = 0.42

b) Two events are said to be independent when occurrence of one event does not affect the probability of the other event's occurrence. Here the person orders Afghani Chicken regardless of where he visits so the events are independent.

c)  P = 0.30 because he orders Afghani Chicken regardless of where he visits.

d)  Let A be the probability that he goes to KARIM:

P(A) = 0.7 * ( 1 - 0.6 ) = 0.28

Let A be the probability that he orders Afghani Chicken:

P(B) =  0.3 * 0.6 = 0.18

Let C be the probability that he goes to KARIM and orders Afghani chicken:

= 0.7 * 0.6 = 0.42

So probability that he goes to KARIM or orders Afghani Chicken or both:

P(A) + P(B) + P(C) = 0.28 + 0.18 + 0.42 = 0.88

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Answer:

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[721] :      

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[110]  :

Here z component is zero.

That is why this is  in x -y plane

From the cubic unit directions we can easily understand all given directions  [012], [721], [110].

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3 years ago
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How to stop toilets  

Explanation:

I think

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Explain how collecting mean values (X-bar) from samples drawn from a dataset, regardless of the statistical distribution of the
Pani-rosa [81]

Answer:

answer is given below

Explanation:

Central Limit Theorem: The Central Limit Theorem (CLT) is a statistical theory that gives a sufficiently large sample size with limited variations from the population, the average of all samples from the same population is approximately the same. . In addition, all models follow a nearly normal distribution model.

The given phenomenon is described in the central limit theory. In other words, if we repeatedly take independent random samples of size n from any population, when n is large, the sample distribution is the normal distribution pattern.

mean of the sample means

                    \mu _\bar x = \mu        .............1

and here standard deviation of the sample means is

                      \mu _\bar x = \frac{\sigma }{\sqrt{n}}      .........2

and This theory is found elsewhere in the field of statistics. Although the central limit theory may seem abstract and devoid of any application, this theory is actually important for statistical practice.

8 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}]

 

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vovangra [49]

Answer:

The correct answer is "O (n\times Log n)". A further explanation is given below.

Explanation:

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