Mathematics : asked on jaelynnm
 26.05.2023

what is the probability of time drawing a two and then a five without replacement from the dealer's hand

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Mathematics
Step-by-step answer
P Answered by Specialist

For probabilities with replacement

P(2\ Red) = \frac{25}{64}

P(2\ Black) = \frac{9}{64}

P(1\ Red\ and\ 1\ Black) = \frac{15}{32}

P(1st\ Red\ and\ 2nd\ Black) = \frac{15}{64}

For probabilities without replacement

P(2\ Red) = \frac{5}{14}

P(2\ Black) = \frac{3}{28}

P(1\ Red\ and\ 1\ Black) = \frac{15}{28}

P(1st\ Red\ and\ 2nd\ Black) = \frac{15}{56}

Step-by-step explanation:

Given

Marbles = 8

Red = 5

Black = 3

For probabilities with replacement

(a) P(2 Red)

This is calculated as:

P(2\ Red) = P(Red)\ and\ P(Red)

P(2\ Red) = P(Red)\ *\ P(Red)

So, we have:

P(2\ Red) = \frac{n(Red)}{Total} \ *\ \frac{n(Red)}{Total}\\

P(2\ Red) = \frac{5}{8} * \frac{5}{8}

P(2\ Red) = \frac{25}{64}

(b) P(2 Black)

This is calculated as:

P(2\ Black) = P(Black)\ and\ P(Black)

P(2\ Black) = P(Black)\ *\ P(Black)

So, we have:

P(2\ Black) = \frac{n(Black)}{Total}\ *\ \frac{n(Black)}{Total}

P(2\ Black) = \frac{3}{8}\ *\ \frac{3}{8}

P(2\ Black) = \frac{9}{64}

(c) P(1 Red and 1 Black)

This is calculated as:

P(1\ Red\ and\ 1\ Black) = [P(Red)\ and\ P(Black)]\ or\ [P(Black)\ and\ P(Red)]

P(1\ Red\ and\ 1\ Black) = [P(Red)\ *\ P(Black)]\ +\ [P(Black)\ *\ P(Red)]

P(1\ Red\ and\ 1\ Black) = 2[P(Red)\ *\ P(Black)]

So, we have:

P(1\ Red\ and\ 1\ Black) = 2*[\frac{5}{8} *\frac{3}{8}]

P(1\ Red\ and\ 1\ Black) = 2*\frac{15}{64}

P(1\ Red\ and\ 1\ Black) = \frac{15}{32}

(d) P(1st Red and 2nd Black)

This is calculated as:

P(1st\ Red\ and\ 2nd\ Black) = [P(Red)\ and\ P(Black)]

P(1st\ Red\ and\ 2nd\ Black) = P(Red)\ *\ P(Black)

P(1st\ Red\ and\ 2nd\ Black) = \frac{n(Red)}{Total}  *\ \frac{n(Black)}{Total}

So, we have:

P(1st\ Red\ and\ 2nd\ Black) = \frac{5}{8} *\frac{3}{8}

P(1st\ Red\ and\ 2nd\ Black) = \frac{15}{64}

For probabilities without replacement

(a) P(2 Red)

This is calculated as:

P(2\ Red) = P(Red)\ and\ P(Red)

P(2\ Red) = P(Red)\ *\ P(Red)

So, we have:

P(2\ Red) = \frac{n(Red)}{Total} \ *\ \frac{n(Red)-1}{Total-1}

We subtracted 1 because the number of red balls (and the total) decreased by 1 after the first red ball is picked.

P(2\ Red) = \frac{5}{8} * \frac{4}{7}

P(2\ Red) = \frac{5}{2} * \frac{1}{7}

P(2\ Red) = \frac{5}{14}

(b) P(2 Black)

This is calculated as:

P(2\ Black) = P(Black)\ and\ P(Black)

P(2\ Black) = P(Black)\ *\ P(Black)

So, we have:

P(2\ Black) = \frac{n(Black)}{Total}\ *\ \frac{n(Black)-1}{Total-1}

We subtracted 1 because the number of black balls (and the total) decreased by 1 after the first black ball is picked.

P(2\ Black) = \frac{3}{8}\ *\ \frac{2}{7}

P(2\ Black) = \frac{3}{4}\ *\ \frac{1}{7}

P(2\ Black) = \frac{3}{28}

(c) P(1 Red and 1 Black)

This is calculated as:

P(1\ Red\ and\ 1\ Black) = [P(Red)\ and\ P(Black)]\ or\ [P(Black)\ and\ P(Red)]

P(1\ Red\ and\ 1\ Black) = [P(Red)\ *\ P(Black)]\ +\ [P(Black)\ *\ P(Red)]

P(1\ Red\ and\ 1\ Black) = [\frac{n(Red)}{Total}\ *\ \frac{n(Black)}{Total-1}]\ +\ [\frac{n(Black)}{Total}\ *\ \frac{n(Red)}{Total-1}]

So, we have:

P(1\ Red\ and\ 1\ Black) = [\frac{5}{8} *\frac{3}{7}] + [\frac{3}{8} *\frac{5}{7}]

P(1\ Red\ and\ 1\ Black) = [\frac{15}{56} ] + [\frac{15}{56}]

P(1\ Red\ and\ 1\ Black) = \frac{30}{56}

P(1\ Red\ and\ 1\ Black) = \frac{15}{28}

(d) P(1st Red and 2nd Black)

This is calculated as:

P(1st\ Red\ and\ 2nd\ Black) = [P(Red)\ and\ P(Black)]

P(1st\ Red\ and\ 2nd\ Black) = P(Red)\ *\ P(Black)

P(1st\ Red\ and\ 2nd\ Black) = \frac{n(Red)}{Total}  *\ \frac{n(Black)}{Total-1}

So, we have:

P(1st\ Red\ and\ 2nd\ Black) = \frac{5}{8} *\frac{3}{7}

P(1st\ Red\ and\ 2nd\ Black) = \frac{15}{56}

Mathematics
Step-by-step answer
P Answered by Master

For probabilities with replacement

P(2\ Red) = \frac{25}{64}

P(2\ Black) = \frac{9}{64}

P(1\ Red\ and\ 1\ Black) = \frac{15}{32}

P(1st\ Red\ and\ 2nd\ Black) = \frac{15}{64}

For probabilities without replacement

P(2\ Red) = \frac{5}{14}

P(2\ Black) = \frac{3}{28}

P(1\ Red\ and\ 1\ Black) = \frac{15}{28}

P(1st\ Red\ and\ 2nd\ Black) = \frac{15}{56}

Step-by-step explanation:

Given

Marbles = 8

Red = 5

Black = 3

For probabilities with replacement

(a) P(2 Red)

This is calculated as:

P(2\ Red) = P(Red)\ and\ P(Red)

P(2\ Red) = P(Red)\ *\ P(Red)

So, we have:

P(2\ Red) = \frac{n(Red)}{Total} \ *\ \frac{n(Red)}{Total}\\

P(2\ Red) = \frac{5}{8} * \frac{5}{8}

P(2\ Red) = \frac{25}{64}

(b) P(2 Black)

This is calculated as:

P(2\ Black) = P(Black)\ and\ P(Black)

P(2\ Black) = P(Black)\ *\ P(Black)

So, we have:

P(2\ Black) = \frac{n(Black)}{Total}\ *\ \frac{n(Black)}{Total}

P(2\ Black) = \frac{3}{8}\ *\ \frac{3}{8}

P(2\ Black) = \frac{9}{64}

(c) P(1 Red and 1 Black)

This is calculated as:

P(1\ Red\ and\ 1\ Black) = [P(Red)\ and\ P(Black)]\ or\ [P(Black)\ and\ P(Red)]

P(1\ Red\ and\ 1\ Black) = [P(Red)\ *\ P(Black)]\ +\ [P(Black)\ *\ P(Red)]

P(1\ Red\ and\ 1\ Black) = 2[P(Red)\ *\ P(Black)]

So, we have:

P(1\ Red\ and\ 1\ Black) = 2*[\frac{5}{8} *\frac{3}{8}]

P(1\ Red\ and\ 1\ Black) = 2*\frac{15}{64}

P(1\ Red\ and\ 1\ Black) = \frac{15}{32}

(d) P(1st Red and 2nd Black)

This is calculated as:

P(1st\ Red\ and\ 2nd\ Black) = [P(Red)\ and\ P(Black)]

P(1st\ Red\ and\ 2nd\ Black) = P(Red)\ *\ P(Black)

P(1st\ Red\ and\ 2nd\ Black) = \frac{n(Red)}{Total}  *\ \frac{n(Black)}{Total}

So, we have:

P(1st\ Red\ and\ 2nd\ Black) = \frac{5}{8} *\frac{3}{8}

P(1st\ Red\ and\ 2nd\ Black) = \frac{15}{64}

For probabilities without replacement

(a) P(2 Red)

This is calculated as:

P(2\ Red) = P(Red)\ and\ P(Red)

P(2\ Red) = P(Red)\ *\ P(Red)

So, we have:

P(2\ Red) = \frac{n(Red)}{Total} \ *\ \frac{n(Red)-1}{Total-1}

We subtracted 1 because the number of red balls (and the total) decreased by 1 after the first red ball is picked.

P(2\ Red) = \frac{5}{8} * \frac{4}{7}

P(2\ Red) = \frac{5}{2} * \frac{1}{7}

P(2\ Red) = \frac{5}{14}

(b) P(2 Black)

This is calculated as:

P(2\ Black) = P(Black)\ and\ P(Black)

P(2\ Black) = P(Black)\ *\ P(Black)

So, we have:

P(2\ Black) = \frac{n(Black)}{Total}\ *\ \frac{n(Black)-1}{Total-1}

We subtracted 1 because the number of black balls (and the total) decreased by 1 after the first black ball is picked.

P(2\ Black) = \frac{3}{8}\ *\ \frac{2}{7}

P(2\ Black) = \frac{3}{4}\ *\ \frac{1}{7}

P(2\ Black) = \frac{3}{28}

(c) P(1 Red and 1 Black)

This is calculated as:

P(1\ Red\ and\ 1\ Black) = [P(Red)\ and\ P(Black)]\ or\ [P(Black)\ and\ P(Red)]

P(1\ Red\ and\ 1\ Black) = [P(Red)\ *\ P(Black)]\ +\ [P(Black)\ *\ P(Red)]

P(1\ Red\ and\ 1\ Black) = [\frac{n(Red)}{Total}\ *\ \frac{n(Black)}{Total-1}]\ +\ [\frac{n(Black)}{Total}\ *\ \frac{n(Red)}{Total-1}]

So, we have:

P(1\ Red\ and\ 1\ Black) = [\frac{5}{8} *\frac{3}{7}] + [\frac{3}{8} *\frac{5}{7}]

P(1\ Red\ and\ 1\ Black) = [\frac{15}{56} ] + [\frac{15}{56}]

P(1\ Red\ and\ 1\ Black) = \frac{30}{56}

P(1\ Red\ and\ 1\ Black) = \frac{15}{28}

(d) P(1st Red and 2nd Black)

This is calculated as:

P(1st\ Red\ and\ 2nd\ Black) = [P(Red)\ and\ P(Black)]

P(1st\ Red\ and\ 2nd\ Black) = P(Red)\ *\ P(Black)

P(1st\ Red\ and\ 2nd\ Black) = \frac{n(Red)}{Total}  *\ \frac{n(Black)}{Total-1}

So, we have:

P(1st\ Red\ and\ 2nd\ Black) = \frac{5}{8} *\frac{3}{7}

P(1st\ Red\ and\ 2nd\ Black) = \frac{15}{56}

Mathematics
Step-by-step answer
P Answered by Specialist

(B) Based on the parents' genetics, each of 6 children from a particular pair of parents has a 0.36 probability of having lactose intolerance. The random variable represents the total number of children from this pair of parents with lactose intolerance.

True we have a binomial variable since we have a value of n defined n =6 and a probability of success 0.36 with the the condition of success and failure.

(C) There are two choices of burritos at a restaurant, vegetarian or beef. The random variable represents the total number out of 254 customers who ordered beef.

True we have a binomial variable since we have a value of n defined n =254 and a probability of success 1/2 since we have just two possible options, and we satisfy the condition of success and failure.

(E) A coin flip has two outcomes: heads or tails. The probability of each outcome is 0.50. The random variable represents the total number of flips required to get tails.

True we have a binomial variable since we have a value of n for the experiment and a probability of success 1/2 since we have just two possible options, and we satisfy the condition of success and failure.

Step-by-step explanation:

Let's analyze one by one the possible options:

(A) A quality check on a particular product must meet five guidelines. All products are made in the same factory under the same conditions. The random variable represents the total number of products out of 35 tested that pass inspection.

False, for this case we don't have a binomial variable because the outcome is not under the criteria of success or failure of an event because we can have more than two criteria to classify the product, and we don't have a probability defined for the success or failure.

(B) Based on the parents' genetics, each of 6 children from a particular pair of parents has a 0.36 probability of having lactose intolerance. The random variable represents the total number of children from this pair of parents with lactose intolerance.

True we have a binomial variable since we have a value of n defined n =6 and a probability of success 0.36 with the the condition of success and failure.

(C) There are two choices of burritos at a restaurant, vegetarian or beef. The random variable represents the total number out of 254 customers who ordered beef.

True we have a binomial variable since we have a value of n defined n =254 and a probability of success 1/2 since we have just two possible options, and we satisfy the condition of success and failure.

(D) The probability of drawing a king in a standard deck of cards is 0.08. Seven cards are drawn without replacement. The random variable represents the total number of king cards observed.

False since the experiment is without replacelente each time that we select a king the probability of select another king is not the same and for this reason we can't have a binomial experiment.

(E) A coin flip has two outcomes: heads or tails. The probability of each outcome is 0.50. The random variable represents the total number of flips required to get tails.

True we have a binomial variable since we have a value of n for the experiment and a probability of success 1/2 since we have just two possible options, and we satisfy the condition of success and failure.

Mathematics
Step-by-step answer
P Answered by PhD

A & C

Step-by-step explanation:

Option A: The variable described is binomial because we have a value of n for the experiment and a probability of success 0.5 due to the fact that we have only two possible outcomes, which satisfy the condition of chances of success and failure.

Option B: The variable described is not binomial because the outcome does not fall under the criteria of chances of success or failure of due to the fact that it's possible we have more than two criteria to classify the product.

Option C: The variable described is binomial because we have a definite value of n = 6 with a probability of success = 0.3. Thus, probability of failure will be 0.7 and so it fulfills the condition of success and failure.

Option D: The variable described is not binomial because we don't know if the 2 options are independent of each other.

Option E: Variable describes is not binomial because the cards are drawn without replacement. Also, each time that king is selected, the probability of another king being selected is not the same and for thus the variable is not a a binomial

Mathematics
Step-by-step answer
P Answered by PhD

a. , c. , d., e.

Step-by-step explanation:

A variable that counts how many times a certain event occurs in a particular number of trials is known as binomial random variable.

For each trial, there exist only two outcomes .

The probability of for each event is the same on each trial.

a.  Event has two outcomes with same probability as 0.50, therefore the  random variable represents the total number of flips required to get tails is a binomial random variable.

b. Total guidelines are 5.

Here total outcomes are not 2 , it does not meet with the conditions of binomial.

c. The random variable represents the total number of children from this pair of parents with blue eyes has two outcomes (where has or not.)

also, the probability of having blue eyes is same in each trial, so it represents binomial random variable.

d. The random variable represents the total number out of 567 customers with a checking account has two outcomes (checking or savings).

So, it represents binomial random variable.

e. The random variable represents the total number of ace cards observed has two outcomes ( ace or not ace).

So it represents the binomial random variable.

Mathematics
Step-by-step answer
P Answered by PhD

a. , c. , d., e.

Step-by-step explanation:

A variable that counts how many times a certain event occurs in a particular number of trials is known as binomial random variable.

For each trial, there exist only two outcomes .

The probability of for each event is the same on each trial.

a.  Event has two outcomes with same probability as 0.50, therefore the  random variable represents the total number of flips required to get tails is a binomial random variable.

b. Total guidelines are 5.

Here total outcomes are not 2 , it does not meet with the conditions of binomial.

c. The random variable represents the total number of children from this pair of parents with blue eyes has two outcomes (where has or not.)

also, the probability of having blue eyes is same in each trial, so it represents binomial random variable.

d. The random variable represents the total number out of 567 customers with a checking account has two outcomes (checking or savings).

So, it represents binomial random variable.

e. The random variable represents the total number of ace cards observed has two outcomes ( ace or not ace).

So it represents the binomial random variable.

Mathematics
Step-by-step answer
P Answered by Specialist

(B) Based on the parents' genetics, each of 6 children from a particular pair of parents has a 0.36 probability of having lactose intolerance. The random variable represents the total number of children from this pair of parents with lactose intolerance.

True we have a binomial variable since we have a value of n defined n =6 and a probability of success 0.36 with the the condition of success and failure.

(C) There are two choices of burritos at a restaurant, vegetarian or beef. The random variable represents the total number out of 254 customers who ordered beef.

True we have a binomial variable since we have a value of n defined n =254 and a probability of success 1/2 since we have just two possible options, and we satisfy the condition of success and failure.

(E) A coin flip has two outcomes: heads or tails. The probability of each outcome is 0.50. The random variable represents the total number of flips required to get tails.

True we have a binomial variable since we have a value of n for the experiment and a probability of success 1/2 since we have just two possible options, and we satisfy the condition of success and failure.

Step-by-step explanation:

Let's analyze one by one the possible options:

(A) A quality check on a particular product must meet five guidelines. All products are made in the same factory under the same conditions. The random variable represents the total number of products out of 35 tested that pass inspection.

False, for this case we don't have a binomial variable because the outcome is not under the criteria of success or failure of an event because we can have more than two criteria to classify the product, and we don't have a probability defined for the success or failure.

(B) Based on the parents' genetics, each of 6 children from a particular pair of parents has a 0.36 probability of having lactose intolerance. The random variable represents the total number of children from this pair of parents with lactose intolerance.

True we have a binomial variable since we have a value of n defined n =6 and a probability of success 0.36 with the the condition of success and failure.

(C) There are two choices of burritos at a restaurant, vegetarian or beef. The random variable represents the total number out of 254 customers who ordered beef.

True we have a binomial variable since we have a value of n defined n =254 and a probability of success 1/2 since we have just two possible options, and we satisfy the condition of success and failure.

(D) The probability of drawing a king in a standard deck of cards is 0.08. Seven cards are drawn without replacement. The random variable represents the total number of king cards observed.

False since the experiment is without replacelente each time that we select a king the probability of select another king is not the same and for this reason we can't have a binomial experiment.

(E) A coin flip has two outcomes: heads or tails. The probability of each outcome is 0.50. The random variable represents the total number of flips required to get tails.

True we have a binomial variable since we have a value of n for the experiment and a probability of success 1/2 since we have just two possible options, and we satisfy the condition of success and failure.

Mathematics
Step-by-step answer
P Answered by PhD

1. 72

2. 216

3a. 2/17

3b. 1/425

Step-by-step explanation:

1.

Let's think about the first one a little bit. To simplify things, let's assume that Jimmy always sits down first, followed by Sean and then everyone else. If Jimmy sits at either end of the row, then Sean has 3 choices, followed by 3 choices for the next person, 2 for the next, and only 1 remaining for the last one. Therefore, we can divide this problem into two cases:

A) Jimmy sits at one of the sides

In this case, there are 2 choices for Jimmy, and then the amount of choices shown above, meaning that there are a total of 2*3*3*2*1=36 ways for them to be seated.

B) Jimmy sits in one of the center seats

In this case, there are two seats to either side of Jimmy that Sean will not sit in, leaving him two options. In this scenario, there are 2 options for Sean and 3 for Jimmy, which means there are a total of 3*2*3*2*1=36 ways for them to be sat.

Adding these two together, you get a total of 72 possible ways.

2.

Since all of the vowels cannot move around each other, we are essentially just finding all of the ways of arranging the consonants inside the word. Since the vowels must be stuck in the order AEIOU, then there are 6 positions for the letter Q, 6 for the letter T, and 6 for the letter N, making a total of 216 possible arrangements.

3a. The probability that the first card is red is 1/2, since half of the cards in the deck are red. Without replacement, the probability that the second card is red if the first one is will be 25/51, and the third card is 24/50. Multiplying these together, you get 25*24/50*51*2=2/17.

3b. It doesn't matter what the rank of the first card is, only whether the other two are the same rank. The probability of the second card being of the same rank as the first is 3/51, and for the third is 2/50, which when multiplied together gives 1/425. Hope this helps!

Mathematics
Step-by-step answer
P Answered by PhD

1. 72

2. 216

3a. 2/17

3b. 1/425

Step-by-step explanation:

1.

Let's think about the first one a little bit. To simplify things, let's assume that Jimmy always sits down first, followed by Sean and then everyone else. If Jimmy sits at either end of the row, then Sean has 3 choices, followed by 3 choices for the next person, 2 for the next, and only 1 remaining for the last one. Therefore, we can divide this problem into two cases:

A) Jimmy sits at one of the sides

In this case, there are 2 choices for Jimmy, and then the amount of choices shown above, meaning that there are a total of 2*3*3*2*1=36 ways for them to be seated.

B) Jimmy sits in one of the center seats

In this case, there are two seats to either side of Jimmy that Sean will not sit in, leaving him two options. In this scenario, there are 2 options for Sean and 3 for Jimmy, which means there are a total of 3*2*3*2*1=36 ways for them to be sat.

Adding these two together, you get a total of 72 possible ways.

2.

Since all of the vowels cannot move around each other, we are essentially just finding all of the ways of arranging the consonants inside the word. Since the vowels must be stuck in the order AEIOU, then there are 6 positions for the letter Q, 6 for the letter T, and 6 for the letter N, making a total of 216 possible arrangements.

3a. The probability that the first card is red is 1/2, since half of the cards in the deck are red. Without replacement, the probability that the second card is red if the first one is will be 25/51, and the third card is 24/50. Multiplying these together, you get 25*24/50*51*2=2/17.

3b. It doesn't matter what the rank of the first card is, only whether the other two are the same rank. The probability of the second card being of the same rank as the first is 3/51, and for the third is 2/50, which when multiplied together gives 1/425. Hope this helps!

Mathematics
Step-by-step answer
P Answered by Specialist

Sampling with replacement.

a) 2/5 = 0.4

b) 8/125 = 0.064

c) 7/200 = 0.035

d) 1127/4000 = 0.28175

Sampling without replacement

a) 2/5 = 0.4

b) 14/285 = 0.0491

c) 7/171 = 0.0409

d) 861/3420 = 0.25175

Step-by-step explanation:

Total number of marbles = 8+7+5 = 20

Probability of drawing a Blue marble = P(B) = 8/20 = 2/5

Probability of drawing a Red marble = P(R) = 7/20

Probability of drawing a green marble = P(G) = 5/20 = 1/4

Sampling with replacement.

a) Probability of a blue marble in one draw from the bag = 8/20 = 2/5 = 0.4

b) Probability of three blue marbles in three draws from the bag = (8/20) × (8/20) × (8/20) = 8/125

c) Probability of a red, a green, and a blue marble in that order in three draws from the bag = (7/20) × (5/20) × (8/20) = 280/8000 = 7/200

d) Probability of at least two red marbles in three draws from the bag

This is a sum of the probability of drawing two red marbles out of three and three red marbles from three draws.

Probability of drawing two red marbles out of three draws = 3 × P(R) × P(R) × P(other balls) = (7/20) × (7/20) × (13/20) = 1911/8000

The 3 is there because this can be done in 3 different orders

P(other balls) = 1 - (7/20) = 13/20

Probability of drawing three red marbles in three draws = (7/20) × (7/20) × (7/20) = 343/8000

Probability of at least two red marbles in three draws from the bag = (1911/8000) + (343/8000) = 2254/8000 = 1127/4000

Sampling without replacement

In this sampling method, the sample space and number of marbles decrease as each marble goes out of the bag.

a) Probability of a blue marble in one draw from the bag = 8/20 = 2/5 = 0.4

b) Probability of three blue marbles in three draws from the bag = (8/20) × (7/19) × (6/18) = 336/6840 = 14/285

c) Probability of a red, a green, and a blue marble in that order in three draws from the bag = (7/20) × (5/19) × (8/18) = 280/6840 = 7/171

d) Probability of at least two red marbles in three draws from the bag

This is a sum of the probability of drawing two red marbles out of three and three red marbles from three draws.

Probability of drawing two red marbles out of three draws = 3 × (7/20) × (6/19) × (12/18) = 1512/6840

The 3 is there to multiply because this can be done in three different orders

Probability of drawing three red marbles in three draws = (7/20) × (6/19) × (5/18) = 210/6840

Probability of at least two red marbles in three draws from the bag = (1512/6840) + (210/6840) = 1722/8000 = 861/3420

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