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Elan Coil [88]
1 year ago
9

I don't know if it's surjective,injective,bijective or none, please help it's due today​

Mathematics
1 answer:
Bond [772]1 year ago
7 0

The given polynomial graph can be classified as; surjective

<h3>How to Interpret Mapping Functions?</h3>

A General Function points from each member of "A" to a member of "B".

It never has one "A" pointing to more than one "B", so one-to-many is not OK in a function

Injective means we won't have two or more "A"s pointing to the same "B".

So many-to-one is not okay (which is Okay for a general function).

As it is also a function one-to-many is not Okay.

Surjective means that every "B" has at least one matching "A". There won't be a "B" left out.

Bijective means both Injective and Surjective together.

It's more like a "perfect pairing" between the sets: every one has a partner and no one is left out.

So there is a perfect "one-to-one correspondence" between the members of the sets.

Looking at the given graph, it is not a one to one as it is clearly surjective.

Read more about Mapping Functions at; brainly.com/question/7382340

#SPJ1

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c= \frac{1}{4}d

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Answer:

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Step-by-step explanation:

Given

f \left(x,y \right) = \left{ \begin{array} { l l } { k , } & { 0 \leq x} \leq 2,0 \leq y \leq 1,2 y  \leq x }  & { \text 0, { elsewhere. } } \end{array} \right.

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Find k

To solve for k, we use the definition of joint probability function:

\int\limits^a_b \int\limits^a_b {f(x,y)} \, = 1

Where

{ 0 \leq x} \leq 2,0 \leq y \leq 1,2 y  \leq x }

Substitute values for the interval of x and y respectively

So, we have:

\int\limits^2_{0} \int\limits^{x/2}_{0} {k\ dy\ dx} \, = 1

Isolate k

k \int\limits^2_{0} \int\limits^{x/2}_{0} {dy\ dx} \, = 1

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k \int\limits^2_{0} y {dx} \, [0,x/2]= 1

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k \int\limits^2_{0} (x/2 - 0) {dx} \,= 1

k \int\limits^2_{0} \frac{x}{2} {dx} \,= 1

Integrate x

k * \frac{x^2}{2*2} [0,2]= 1

k * \frac{x^2}{4} [0,2]= 1

Substitute 0 and 2 for x

k *[ \frac{2^2}{4} - \frac{0^2}{4} ]= 1

k *[ \frac{4}{4} - \frac{0}{4} ]= 1

k *[ 1-0 ]= 1

k *[ 1]= 1

k = 1

Solving (b): P(x > 3y)

We have:

f(x,y) = k

Where k = 1

f(x,y) = 1

To find P(x > 3y), we use:

\int\limits^a_b \int\limits^a_b {f(x,y)}

So, we have:

P(x > 3y) = \int\limits^2_0 \int\limits^{y/3}_0 {f(x,y)} dxdy

P(x > 3y) = \int\limits^2_0 \int\limits^{y/3}_0 {1} dxdy

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Substitute 0 and y/3 for x

P(x > 3y) = \int\limits^2_0  [y/3 - 0]dy

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Integrate

P(x > 3y) = \frac{y^2}{2*3} [0,2]

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Substitute 0 and 2 for y

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P(x > 3y) = \frac{4}{6} -\frac{0}{6}

P(x > 3y) = \frac{4}{6}

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