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2 years ago

Equation for this slope

Mathematics

1 answer:

spin [16.1K]2 years ago

5 0

Answer:

m = y1 - y2/ x1 -x2

Step-by-step explanation:

Your coordinates are (4, 1) and (0, -3) so m (slope) equals 1 - (-3) over 4 - 0 = 4/4 = 1

Hope this helped!

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In a small chess tournament, 20 matches were played. Find out how many people were involved if it is known that each participant

Nina [5.8K]

Answer:

<em>5</em><em> players participated in the tournament.</em>

Step-by-step explanation:

In a small chess tournament, 20 matches were played.

Let us assume that n number of players participated in the tournament

As in each game 2 players play, so the number of ways they can play is,

$=\ ^nC_2$

As they played 2 games with every other participant in the tournament.

So the total number of games is,

$=\ 2\times ^nC_2$

But it is given to be 20, so

$\Rightarrow \ 2\times ^nC_2=20$

$\Rightarrow \ ^nC_2=10$

$\Rightarrow \dfrac{n!}{2!(n-2)!}=10$

$\Rightarrow \dfrac{n(n-1)}{2}=10$

$\Rightarrow {n(n-1)=20$

As $5\times 4=20$ , so we get n=5.

Therefore, 5 players participated in the tournament.

5 0

2 years ago

Read 2 more answers

By what percent would the area of the rectangle change if the width of the rectangle is increased by 50% and the length is incre

belka [17]

Answer:

. i think it's wrong

7 0

2 years ago

What is the equation of the asymptote of the graph of y=2(6^x) ?

Arlecino [84]

The value of can never be 0

y = 0

A

6 0

2 years ago

Find the least number which when divisible by 20, 24, 32 and 38 leaves a remainder 5 in cach case

Olegator [25]

Answer:

18245

Step-by-step explanation:

We have to use L.C.M,

L.C.M(20,24,32,38)

2|20,24,32,38

2| 10 ,12 ,16 ,19

2| 5 , 6 , 8 , 19

2| 5 , 3 , 4 , 19

2| 5 , 3 , 2 , 19

L.C.M = 2 x 2 x 2 x 2 x 2 x 2 x 5 x 3 x 19

= 18240

Now for each case remainder is 5,

So the number is 18240+5

=> 18245

7 0

2 years ago

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

3 years ago