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Ksju [112]
3 years ago
6

David wants to rent a movie. He wants to watch either a comedy or a drama. The movie rental store has 18 comedies and dramas ava

ilable for rent. Seven of the movies are comedies, and eleven of the movies are dramas. David has not seen two of the comedies, and he has not seen four of the dramas. If David selects a movie randomly, what is the probability the movie will be a comedy or a movie that he has not seen?
Mathematics
2 answers:
Illusion [34]3 years ago
7 0
Let
A = number of ways to pick a movie David wants (either a comedy or a movie he hasn't seen)

B = number of movies total

There are 7 comedies and 4 dramas that David hasn't seen (note: the comedy movies that he hasn't seen are part of the "comedy movie" count, so there's no need to double count these). So there are 7+4 = 11 movies that David would want to rent. Therefore, A = 11.

There are 18 movies total for rent, so B = 18

Dividing the two values
A/B = 11/18 = 0.6111 approximately

The answer as a fraction is exactly 11/18
The answer in decimal form is roughly 0.6111
The answer in percent form is approximately 61.11%
chubhunter [2.5K]3 years ago
3 0

|\Omega|=18\\ |A|=7+4=11\\\\ P(A)=\dfrac{11}{18}\approx61\%

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You can use the fact that number of breads purchased cannot be negative since a customer either buys them or not and usually do not sell to the shopkeeper.(if somehow they end up selling to shop owner, then yes that will go in negative, but we'll assume it is wrong in most cases as generally shop owners are there to sell stuffs).

The third table of values matches the equation and includes only viable solutions.

<h3>What is a viable solution here?</h3>

It is talking about those solutions which are seen in real world. As stated above, a customer either buys the bread or not, thus number of breads sold will be either positive or 0(in case of no selling). Thus, we cannot have number of breads as negative.

Such solutions which are correct in the real world context here are called here as viable solutions.

<h3>Checking one by one all the tables for them being matched with table and viability</h3>

For first table, the number of breads are in negative, thus it is not going to have viable solution.

For second table, we have:

b = 0 thus c = 3.5b = 3.5 times 0 = 0 which is correctly given in second column.

b = 0.5, thus c = 3.5b = 3.5 times 0.5 =1.75 which is correctly given.

b = 1, thus c= 3.5 times 1 = 3.5 which is correctly given

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For third table, we have:

b = 0, thus c = 3.5 \times 0 = 0, correctly given in second column.

b = 3, thus c = 3.5 \times 3 = 10.5, correctly given.

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b = 9, thus c = 3.5 \times 9 = 31.5, correctly given.

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Learn more about purchasing to cost relation here:
brainly.com/question/13727919

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xz_007 [3.2K]

Answer:

 \large\boxed{\large\boxed{2.428678219969\times 10^{15}}}

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It is said that the <em>order is not considered</em>, meaning that the order does not matter. For instance, ABCDEF is equal to any combination of those same letters.

Thus, this problem deals with combinations, whose formula is:

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For <em>27 flavors of ice cream</em> and <em>6 dip sundaes</em>, m = 27 and n = 6:

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Step-by-step explanation:

6/8 = 0.75

Or 8/6 = 1.3 repeating

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