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3 years ago

Which of the following relations is not a function? Select all.

Mathematics

1 answer:

elena-14-01-66 [18.8K]3 years ago

6 0

Hi there!

A function is a set of ordered pairs that do not have same x-values.

In a set of ordered pairs, If two or more ordered pairs have same x-values, they are not considered as a function but rather a relation instead.

From the question above, you notice that two of the choices have same x-values and others don't have.

The relations that are not functions are the <u>second choice and fourth choice</u> because two ordered pairs have same x-values.

Any questions can be asked through comment.

Furthermore, a function is also a relation but a relation isn't always a function.

Hope this helps, and Happy Learning! :)

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(CO 3) Fifty-four percent of US teens have heard of a fax machine. You randomly select 12 US teens. Find the probability that th

11111nata11111 [884]

Answer:

a) $P(X=6)=(12C6)(0.54)^6 (1-0.54)^{12-6}=0.217$

b) $P(X> 8) = P(X\geq 9)= P(X=9)+P(X=10)+P(X=11)+P(X=12)$

$P(X=9)=(12C9)(0.54)^9 (1-0.54)^{12-9}=0.0836$

$P(X=10)=(12C10)(0.54)^{10} (1-0.54)^{12-10}=0.0294$

$P(X=11)=(12C11)(0.54)^{11} (1-0.54)^{12-11}=0.00628$

$P(X=12)=(12C12)(0.54)^{12} (1-0.54)^{12-12}=0.000615$

And adding the values we got:

$P(X> 8) = P(X\geq 9)= P(X=9)+P(X=10)+P(X=11)+P(X=12)= 0.0836+0.0294+0.00628+0.000615 = 0.120$

Step-by-step explanation:

Previous concepts

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".

Solution to the problem

Let X the random variable of interest "number of teens that have heard of a fax machine", on this case we now that:

$X \sim Binom(n=12, p=0.54)$

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:

$nCx=\frac{n!}{(n-x)! x!}$

Part a

For this case we want this probability:

$P(X=6)=(12C6)(0.54)^6 (1-0.54)^{12-6}=0.217$

Part b

For this case we want this probability:

$P(X> 8) = P(X\geq 9)= P(X=9)+P(X=10)+P(X=11)+P(X=12)$

$P(X=9)=(12C9)(0.54)^9 (1-0.54)^{12-9}=0.0836$

$P(X=10)=(12C10)(0.54)^{10} (1-0.54)^{12-10}=0.0294$

$P(X=11)=(12C11)(0.54)^{11} (1-0.54)^{12-11}=0.00628$

$P(X=12)=(12C12)(0.54)^{12} (1-0.54)^{12-12}=0.000615$

And adding the values we got:

$P(X> 8) = P(X\geq 9)= P(X=9)+P(X=10)+P(X=11)+P(X=12)= 0.0836+0.0294+0.00628+0.000615 = 0.120$

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