X^2+y^2=1 yes or no relation represents a function or not

Mathematics

1 answer:

sleet_krkn [62]1 year ago

6 0

We are asked to decide if the expression:

x^2 + y^2 = 1 represents a function.

We recall that in order to have a function, we need for a given value of x to have a SINGLE value of y associated with it.

So in this case, when x is 0 for example, we have the following:

0^2 + y^2 = 1

then y^2 = 1

and we realize that there are TWO values of y whose square form gives 1 (one is 1 and the other -1) Therefore, this relationship is NOT a function, since for example when x = 0 there are TWO values of y to which that x is associated (y = 1 and y = -1).

So please select that this is NOT a function for your answer.

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postnew [5]

Answer:

am not sure but

Step-by-step explanation:

since they are similar they must have similar ratio

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so 3:1

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3 years ago

An equation is shown below:

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3 years ago

A new phone-answering system installed by a certain utility company is capable of handling twelve calls every 5 minutes. Prior

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Answer:

A. P(x>12 in 5 minutes)=0.0201

Step-by-step explanation:

Because we are working with a Poisson Distribution of probability, we need to get all the data needed. In a Poisson distribution is needed a constant called λ that symbolizes the mean data (6 calls) per unit of time (5 minutes), for this distribution λ=6/5.

Poisson probabilities work like this:

$P(x=y\ 'in\ z\ unit\ of\ time')= \frac{e^{-z\lambda}(z\lambda)^{y}}{y!}$

Remember y has to be an integer and the units of z must be the same unit of time used in λ. Now we are ready to solve this problem

A. The question is asking for the probability that in 5 minutes appear more calls than the phone-answering machine could answer (i.e P(x>12 in 5 minutes)). Because there are infinite numbers greater than 12, we are using this property of probabilities that´ll help us simplify the problem:

P(x>12 in 5 minutes)= 1 - P(x≤12 in 5 minutes)

Now we can use the following:

$P(x\geq12\ in\ 5\ minutes)=P(x=0\ in\ 5\ min)+P(x=1\ in\ 5\ min)+...P(x=12\ in\ 5\ min)$