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3 years ago

How to write an equation for functions

2 answers:

6 0

Functions main problem is to find the missing variable. A function is an equation in which all of the variables are dependent. For example: if you buy 2 bags of oranges and you pay ten dollars for them but the amount of money you pay for the oranges would change each time you add another bag of oranges and how much you owe for the oranges is the dependent variable. So writing a equation would mainly look like this:

$2x = 10\\v2 = c\\$

<span>So each time you add another bag of oranges the number will increase v3, v4, etc...
</span>

Hope this helps!

Brums [2.3K]3 years ago

5 0

A function is an equation that has only one answer for y for every x. A function assigns exactly one output to each input of a specified type. It is common to name a function either f(x) or g(x) instead of y. f(2) means that we should find the value of our function when x equals 2.

AND KAN U HELP ON MY QUESTION PLEASEE

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Step-by-step explanation:

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2 years ago

Read 2 more answers

Consider writing onto a computer disk and then sending it through a certifier that counts the number of missing pulses. Suppose

Furkat [3]

Answer:

a) 0.164 = 16.4% probability that a disk has exactly one missing pulse

b) 0.017 = 1.7% probability that a disk has at least two missing pulses

c) 0.671 = 67.1% probability that neither contains a missing pulse

Step-by-step explanation:

To solve this question, we need to understand the Poisson distribution and the binomial distribution(for item c).

Poisson distribution:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}
$

In which

x is the number of sucesses

$
e = 2.71828$ is the Euler number

$\mu$ is the mean in the given interval.

Binomial distribution:

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

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

And p is the probability of X happening.

Poisson mean:

$\mu = 0.2$

a. What is the probability that a disk has exactly one missing pulse?

One disk, so Poisson.

This is P(X = 1).

$P(X = 1) = \frac{e^{-0.2}*0.2^{1}}{(1)!} = 0.164
$

0.164 = 16.4% probability that a disk has exactly one missing pulse

b. What is the probability that a disk has at least two missing pulses?

$P(X \geq 2) = 1 - P(X < 2)$

In which

$P(X < 2) = P(X = 0) + P(X = 1)$

In which

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}
$

$P(X = 0) = \frac{e^{-0.2}*0.2^{0}}{(0)!} = 0.819$

$P(X = 1) = \frac{e^{-0.2}*0.2^{1}}{(1)!} = 0.164
$

$P(X < 2) = P(X = 0) + P(X = 1) = 0.819 + 0.164 = 0.983$

$P(X \geq 2) = 1 - P(X < 2) = 1 - 0.983 = 0.017$

0.017 = 1.7% probability that a disk has at least two missing pulses

c. If two disks are independently selected, what is the probability that neither contains a missing pulse?

Two disks, so binomial with n = 2.

A disk has a 0.819 probability of containing no missing pulse, and a 1 - 0.819 = 0.181 probability of containing a missing pulse, so $p = 0.181$

We want to find P(X = 0).

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{2,0}.(0.181)^{0}.(0.819)^{2} = 0.671$

0.671 = 67.1% probability that neither contains a missing pulse

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3 years ago