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3 years ago

Of the following fractions. which is least? 4/7 2/3 3/8 3/5

Mathematics

2 answers:

Karolina [17]3 years ago

7 0

3/8 is your answer, it is .375

4/7 would be a lil over .50

2/3 would be .66*

3/5 would be .60

sorry lol i love fractions hope this helped

Otrada [13]3 years ago

7 0

In order to find the fraction with the least value we need to make sure they have the same denominator. The LCD all these fractions have is 840.

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First fraction.

$\displaystyle\frac{4}{7}$ × $\displaystyle\frac{120}{120} = \displaystyle\frac{480}{840}$

Second fraction.

$\displaystyle\frac{2}{3}$ × $\displaystyle\frac{280}{280}= \displaystyle\frac{560}{840}$

Third fraction.

$\displaystyle\frac{3}{8}$ × $\displaystyle\frac{105}{105} = \displaystyle\frac{315}{840}$

Fourth fraction.

$\displaystyle\frac{3}{5}$ × $\displaystyle\frac{168}{168} = \displaystyle\frac{504}{840}$

According to the converted fractions, the least fraction in the sequence is...

$\displaystyle\frac{3}{8}$

•

- <em> Marlon Nunez</em>

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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

3 years ago

What is the sum of an infinite geometric series if the first term is 81 and the common ratio is 2⁄3?

Marrrta [24]

the sum to infinity of a geometric series = $\frac{a}{1-r}$

where a is the first term and r the common ratio

= $\frac{81}{\frac{1}{3} }$ = 81 × 3 = 243

4 0

3 years ago