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Nastasia [14]
3 years ago
15

What are the coordinates of point S? x $ Mark this and return Next Submit

Mathematics
1 answer:
Galina-37 [17]3 years ago
6 0

Answer:

DONT ASK ME

Step-by-step explanation:

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The probability that two people have the same birthday in a room of 20 people is about 41.1%. It turns out that
salantis [7]

Answer:

a) Let X the random variable of interest, on this case we know that:

X \sim Binom(n=20, p=0.411)

This random variable represent that two people have the same birthday in just one classroom

b) We can find first the probability that one or more pairs of people share a birthday in ONE class. And we can do this:

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

And we can find the individual probability:

P(X=0) = (20C0) (0.411)^0 (1-0.411)^{20-0}=0.0000253

And then:

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

And since we want the probability in the 3 classes we can assume independence and we got:

P= 0.99997^3 = 0.9992

So then the probability that one or more pairs of people share a birthday in your three classes is approximately 0.9992

Step-by-step explanation:

Previous concepts

A Bernoulli trial is "a random experiment with exactly two possible outcomes, "success" and "failure", in which the probability of success is the same every time the experiment is conducted". And this experiment is a particular case of the binomial experiment.

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

The probability mass function for the Binomial distribution is given as:  

P(X)=(nCx)(p)^x (1-p)^{n-x}  

Where (nCx) means combinatory and it's given by this formula:  

nCx=\frac{n!}{(n-x)! x!}  

Solution to the problem

Part a

Let X the random variable of interest, on this case we know that:

X \sim Binom(n=20, p=0.411)

This random variable represent that two people have the same birthday in just one classroom

Part b

We can find first the probability that one or more pairs of people share a birthday in ONE class. And we can do this:

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

And we can find the individual probability:

P(X=0) = (20C0) (0.411)^0 (1-0.411)^{20-0}=0.0000253

And then:

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

And since we want the probability in the 3 classes we can assume independence and we got:

P= 0.99997^3 = 0.9992

So then the probability that one or more pairs of people share a birthday in your three classes is approximately 0.9992

4 0
4 years ago
Help please :) <br> V74 <br> V74
Lelechka [254]

Answer:

x =10

Step-by-step explanation:

Using the pythagoras theorem om one of the right triangle;

(√74)² = (x/2)²+7²

74 = (x/2)²+49

(x/2)² = 74 - 49

(x/2)² = 25

Square root both sides

√ (x/2)² = ±√25

x/2 = ±5

x = 2 * ±5

x = ±10

Hence the value of x is 10

8 0
3 years ago
The corresponding lengths of two cuboids are 12 and 3. What is the ratio of their volumes?
damaskus [11]

Answer:

I think 4:1 I'm not that sure tho

3 0
3 years ago
X = y 2 - 1
pochemuha
Your answer would be option B. 2y² - y - 6 = 0. This is because if you were to substitute x = y² - 1 into the equation 2x - y = 4, you would get 2(y² - 1) - y = 4, which expands into 2y² - 2 - y = 4, and then simplifies to 2y² - y - 6 = 0.
I hope this helps!
7 0
3 years ago
Read 2 more answers
What is the equation for 87÷75.69
vova2212 [387]

Step-by-step explanation:

hope it help you.

thanks.��

5 0
3 years ago
Read 2 more answers
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