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zysi [14]
3 years ago
10

Shane returned from Las Vegas with $700, which is 75% more than what he had in the beginning of the trip, how kuch money did Sha

ne have in the beginning
Mathematics
2 answers:
rjkz [21]3 years ago
7 0
Find 10% of 700 by removing a 0. 10% of 700 is 70. To get 70%, multiply 70 by 10 to get 490. To get 5% of 700, divide 10% (70) by 2 to get 35. Add 35 to 490 to get 75% of 700 (525).

75% of 700 is 525

Now subtract 525 from 700 and you have your answer

175

(This is an easy way to do it in your head)
Ksenya-84 [330]3 years ago
5 0
Well you take 700 * 1.75 = 1,225

so your answer is $1,225
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the name Joe is very common at a school in one out of every ten students go by the name. If there are 15 students in one class,
kumpel [21]

Using the binomial distribution, it is found that there is a 0.7941 = 79.41% probability that at least one of them is named Joe.

For each student, there are only two possible outcomes, either they are named Joe, or they are not. The probability of a student being named Joe is independent of any other student, hence, the <em>binomial distribution</em> is used to solve this question.

<h3>Binomial probability distribution </h3>

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

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

The parameters are:

  • x is the number of successes.
  • n is the number of trials.
  • p is the probability of a success on a single trial.

In this problem:

  • One in ten students are named Joe, hence p = \frac{1}{10} = 0.1.
  • There are 15 students in the class, hence n = 15.

The probability that at least one of them is named Joe is:

P(X \geq 1) = 1 - P(X = 0)

In which:

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

P(X = 0) = C_{15,0}.(0.1)^{0}.(0.9)^{15} = 0.2059

Then:

P(X \geq 1) = 1 - P(X = 0) = 1 - 0.2059 = 0.7941

0.7941 = 79.41% probability that at least one of them is named Joe.

To learn more about the binomial distribution, you can take a look at brainly.com/question/24863377

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