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saul85 [17]
2 years ago
5

Which relation below is NOT a function?

Mathematics
1 answer:
KiRa [710]2 years ago
5 0
Functions have a unique value of y for each value of x.
D has two possible values of y when x = 1 therefore it is NOT a function
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HELP! What is the range of values for X? WILL GIVE BRAINLIEST!
andreev551 [17]

Answer:

Step-by-step explanation:

The range of a function is the set of all possible values it can produce.

For example, consider the function

f(x)=x^{2}

No matter what value we give to x, the function is always positive:

   If x is 2, then the function returns x squared or 4.

   If x is negative 2, then it still produces 4 since -2 times -2 is positive 4.

So the range of the function is

"all real numbers greater than or equal to zero".

7 0
4 years ago
Solve the equation -3 = 8x - 4yfor y
Anestetic [448]

You would add 4y to both sides of the equation which would leave you with

4y - 3 = 8x

then you would divide the entire equation by 4 leaving you with

y - 3/4 = 2x

then you would add 3/4 to both sides of the equation leaving you with

y = 2x + 3/4

8 0
4 years ago
Suppose that we want to generate the outcome of the flip of a fair coin, but that all we have at our disposal is a biased coin w
Goshia [24]

Answer:

Step-by-step explanation:

Given that;

the following procedure for accomplishing our task are:

1. Flip the coin.

2. Flip the coin again.

From here will know that the coin is first flipped twice

3. If both flips land on heads or both land on tails, it implies that we return to step 1 to start again. this makes the flip to be insignificant since both flips land on heads or both land on tails

But if the outcomes of the two flip are different i.e they did not land on both heads or both did not land on tails , then we will consider such an outcome.

Let the probability of head = p

so P(head) = p

the probability of tail be = (1 - p)

This kind of probability follows a conditional distribution and the probability  of getting heads is :

P( \{Tails, Heads\})|\{Tails, Heads,( Heads ,Tails)\})

= \dfrac{P( \{Tails, Heads\})  \cap \{Tails, Heads,( Heads ,Tails)\})}{  {P( \{Tails, Heads,( Heads ,Tails)\}}}

= \dfrac{P( \{Tails, Heads\}) }{  {P( \{Tails, Heads,( Heads ,Tails)\}}}

= \dfrac{P( \{Tails, Heads\}) } {  {P( Tails, Heads) +P( Heads ,Tails)}}

=\dfrac{(1-p)*p}{(1-p)*p+p*(1-p)}

=\dfrac{(1-p)*p}{2(1-p)*p}

=\dfrac{1}{2}

Thus; the probability of getting heads is \dfrac{1}{2} which typically implies that the coin is fair

(b) Could we use a simpler procedure that continues to flip the coin until the last two flips are different and then lets the result be the outcome of the final flip?

For a fair coin (0<p<1) , it's certain that both heads and tails at the end of the flip.

The procedure that is talked about in (b) illustrates that the procedure gives head if and only if the first flip comes out tail with probability 1 - p.

Likewise , the procedure gives tail if and and only if the first flip comes out head with probability of  p.

In essence, NO, procedure (b) does not give a fair coin flip outcome.

5 0
3 years ago
SOMEONE PLEASE ANSWER WITH WORK I WILL GOVE BRAINLIEST
Ira Lisetskai [31]

Answer: what grade are you in

<h3 />

Step-by-step explanation:

5 0
3 years ago
Would you rather have a kilogram of quarters or a kilogram of dimes? Why?
True [87]

Answer:kilogram of quarters

Step-by-step explanation:

because quarters equal 25 cents each and it’s a kilogram

8 0
3 years ago
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