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3 years ago

Which set of rational numbers is arranged from least to greatest?

Mathematics

2 answers:

andrew11 [14]3 years ago

6 0

Answer:

yo goba goba

Step-by-step explanation:

fSA:G

GGGWWGWGWGWg take bathroom break

Aleksandr-060686 [28]3 years ago

3 0

Answer:

-0.5, negative 1 over 5, 1 over 9, 1

Step-by-step explanation:

Hope this helps! :) please mark me as brainly :)

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(7x2) x 3

Sindrei [870]

First you need to multiply the one in the bracket 7x2 which is 14 then u must multiply it with 3 to get the answer so 14x3 which equals to 42

7 0

1 year ago

Find the total surface area of this cone.<br> leave your answer in terms of pi<br> R=3<br> L=4

Yuliya22 [10]

Answer:

total surface area of cone is πr (r+l)

22/7 x 3 (3+4)

22/7*3 (7)

22 x 3 ( as 7 that was divided by 22 was cut out with 7 in the bracket)

hence,

TSA of cone = 66pi

8 0

3 years ago

Write a real word problem that can be represented by the expression 5x+10

trasher [3.6K]

A member ship at the gym cost 5 dollars a month plus a locker holding fee of 10 dollars

3 0

3 years ago

Read 2 more answers

Solve for x: −2x + 5 < 7 <br> Please Help Meh!! DO you LIKE to see people STRUGGLE?!?!

Arlecino [84]

x > -1

-2x+5 < 7

-2x<2

Now since you divide by (-1) to get positive 2,you need to flip the sign

(-2)/(-1)= 2 2/(-1)=-2

2x>-2

x > -1

To check the solution do not replace the x by -1 because it shouls be more than -1 (You can choose 0,1,2,3....)

Example choose x=3

-2x3+5<7

-6+5<7

-1<7

8 0

3 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