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3 years ago

What’s 8.10+12 honest answer please

Mathematics

2 answers:

Alik [6]3 years ago

3 0

The answer is 20.1.

<em>**Please mark this answer as Brainliest and leave a Thanks if I helped :)</em>

DerKrebs [107]3 years ago

3 0

The answer is 20.1. (Hope this helped)

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Three irrational numbers are given below. Select the option that correctly plots the approximate value of each me on live where

Murljashka [212]

The answer is that you are look for is 2

4 0

3 years ago

What is the coefficient of the x5y5-term in the binomial expansion of (2x – 3y)10? 10C5(2)5(3)5 10C5(2)5(–3)5 –10C5(2)5(–3)5 10C

butalik [34]

ANSWER

$10C_5(2)^{5}( - 3)^5$

EXPLANATION

The given binomial expansion is:

${(2x - 3y)}^{10}$

Compare this to

${(a + b)}^{n}$

we have a=2x , b=-3y and n=10

We want to find the coefficient of the term

${x}^{5} {y}^{5}$

This implies that, r=5.

The terms in the expansion can be obtained using

$T_{r+1}=nC_ra^{n-r}b^r$

We substitute the given values to obtain;

$T_{5+1}=10C_5(2x)^{10-5}( - 3y)^5$

$T_{6}=10C_5(2x)^{5}(3y)^5$

$T_{6}=10C_5(2)^{5}( - 3)^5 {x}^{5} {y}^{5}$

Hence the coefficient is;

$10C_5(2)^{5}( - 3)^5$

5 0

3 years ago

Read 2 more answers

Item 3 is unpinned. Click to pin.

Ratling [72]

Answer:

84%

Step-by-step explanation:

i am so sorry if i am wrong. pls no hate tho

7 0

3 years ago

Read 2 more answers

5-2.1=? 5-2.17=? 5-2 7/8 solve make sure to re-check your self before answering or no forty points

frozen [14]

Answer: 5-2.1= 2.9

5-2.17= 2.83

5-2 7/8= 2.125

6 0

4 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

4 years ago