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babunello [35]
2 years ago
6

Which of these situations fit the conditions for using Bernoulli​ trials? Explain. ​a) You are rolling 66 dice and need to get a

t least fourfour 33s to win the game. ​ b) We record the distribution of home stateshome states of customers visiting our website. ​ c) A committee consisting of 1212 men and 88 women selects a delegation of 44 to attend a professional meeting at random. What is the probability they choose all​ women? ​ d) A study found that 5757​% of M.B.A. students admit to cheating. A business school dean surveys all the students in the graduating class and gets responses in which cheating was admitted by 354354 of 542542 students.
Mathematics
1 answer:
shutvik [7]2 years ago
7 0

Answer:

Experiments a) and d) fit the conditions for using Bernoulli trials.

Step-by-step explanation:

A Bernoulli trial is one where the variable is random and dichotomic, that is, it only has two possible outcomes, True/Sucess/Yes/etc. or False/Failure/No/etc. Also, each experiment has the same probability of sucess than the one before and the one after, that means, they are independent. This probability can be calculated by dividing the number of sucess cases by the number of total cases.

Experiment a), where you need four 3s is a Bernoulli trial, as getting a 3 is sucess and not getting a 3 is a failure, and each roll of the dice is independent from each other.

Experiment b) is not a Bernoulli trial as they are more than 2 possible outcomes for the home state of the customer (50 in the case of the US).

Experiment c) is not a Bernoulli trial, as they will be chosen at random, but the first woman will have different chances to be chosen than the fourth one (if they are 20 people, the first one will have 1/20 and the fourth 1/17, as one can't be chosen more than one time).

Experiment d) is a Bernoulli trial, as a student either admits cheating or not, and we can assume that every response was independent from each other.

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Stella [2.4K]

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$12 per hour

Step-by-step explanation:

On the graph it shows he earned $24 in 2 hours, so divide how much he earned by two to get the amount he earned in just one.

7 0
2 years ago
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Use the principle of inclusion and exclusion to find the number of positive integers less than 1,000,000 that are not divisable
wel
This is a simple problem based on combinatorics which can be easily tackled by using inclusion-exclusion principle.
We are asked to find number of positive integers less than 1,000,000 that are not divisible by 6 or 4.
let n be the number of positive integers.
∴ 1≤n≤999,999
Let c₁ be the set of numbers divisible by 6 and c₂ be the set of numbers divisible by 4.
Let N(c₁) be the number of elements in set c₁ and N(c₂) be the number of elements in set c₂.

∴N(c₁) = \frac{999,999}{6} = 166666
N(c₂) = \frac{999,999}{4} = 250000
∴N(c₁c₂) = \frac{999,999}{24} = 41667
∴ Number of positive integers that are not divisible by 4 or 6,

N(c₁`c₂`) = 999,999 - (166666+250000) + 41667 = 625000
Therefore, 625000 integers are not divisible by 6 or 4
8 0
3 years ago
The age of a father is to less than seven times the age of his son in three years the sum oftheir ages will be 52 if the sons pr
vekshin1

Answer:

The correct answer is:

(7s-2)+3+(s+3) = 52, or 8s+4 = 52.

Step-by-step explanation:

Since s is the son's age, "two less than seven times" the son's age would be represented by 7s-2. To represent this in 3 years, we would add 3: (7s-2)+3. In 3 years, the son's age, s, would be represented by s+3. We are told that the sum of these ages will be 52; this gives us (7s-2)+3+(s+3) = 52.

To simplify this, combine like terms. 7s+s = 8s; -2+3+3 = 4. This gives us 8s+4=52.

3 0
3 years ago
Suppose two $20 bills, three $10 bills, one $5 bill, and seven $1 bills are placed in a bag. If you were to pull a bill at rando
Korolek [52]

Answer:

Assuming that the $1 bill was pulled at random, then the expected value of the amount chosen is \frac{7}{13}.

Step-by-step explanation:

From the given question, the bag contains;

$1 bill = 7

$5 bill = 1

$10 bill = 3

$20 bill = 2

Total number of bills in the bag = 13

Pulling a bill at random, the bills would have an expected value as follows:

For $1 bill, the expected value = \frac{7}{13}

For $5 bill, expected value = \frac{1}{13}

For $10 bill, expected value = \frac{3}{13}

For $20 bill, the expected value = \frac{2}{13}

Assuming that the $1 bill was pulled at random, then the expected value of the amount chosen is \frac{7}{13}.

6 0
3 years ago
Type an equation for the following pattern
agasfer [191]

Answer:

y = x + 2

Step-by-step explanation:

It looks like its adding 2 everytime.

y = x + 2

7 0
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