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Evgen [1.6K]
4 years ago
8

Jane and Lee had dinner at The Palace The bill totaled $20 30 with tax The service

Mathematics
1 answer:
kirza4 [7]4 years ago
3 0

Answer:

$3.45

Step-by-step explanation:

$20.30 / 10  = $2.30 = 10%

$2.30 / 2  = $1.15 = 5%

$2.30 + $1.15 = $3.45 = 15%

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which ones do you need answered? all of them? want to help

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Drag the tiles to the table. Each tile can be used more than once. Determine the most likely action that an investor will take f
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Considering the market options that an investor takes, Each scenario listed is attached to the right action that is shown on the tiles:

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<h3>Meaning of an Investor</h3>

An investor can be defined as a name given to an individual that takes every opportunity opened to him to make a profit and by so doing doesn't use his money but puts it into a venture that will yield more.

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8 0
2 years ago
"What is the probability that in 10 dice throws, you will throw AT LEAST two ‘3’s on the dice? Assume you’re throwing a single d
dem82 [27]

Answer:

There is a 51.61% probability that in 10 dice throws, you will throw AT LEAST two ‘3’s on the dice.

Step-by-step explanation:

For each throw, there are only two possible outcomes. Either it is a '3', or it is not. This means that we solve this problem using the binomial probability distribution.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinatios of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

In this problem we have that:

There are 6 possible outcomes for the dice. This means that the probability that it is a '3' is \frac{1}{6} = 0.167

There are 10 throws, so n = 10.

Probability of throwing AT LEAST two ‘3’s on the dice?

Either you throw less than two, or you throw at least two. The sum of the probabilities of these events is 1. So

P(X < 2) + P(X \geq 2) = 1

P(X \geq 2) = 1 - P(X < 2).

In which

P(X < 2) = P(X = 0) + P(X = 1).

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 0) = C_{10,0}.(0.167)^{0}.(0.833)^{10} = 0.1609

P(X = 1) = C_{10,1}.(0.167)^{1}.(0.833)^{9} = 0.3225

So

P(X < 2) = P(X = 0) + P(X = 1) = 0.1609 + 0.3225 = 0.4834.

Finally:

P(X \geq 2) = 1 - P(X < 2) = 1 - 0.4839 = 0.5161.

There is a 51.61% probability that in 10 dice throws, you will throw AT LEAST two ‘3’s on the dice.

5 0
3 years ago
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