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3 years ago

State which of the following sets of ordered pairs represent a function.

Mathematics

2 answers:

anastassius [24]3 years ago

8 0

Answer:

d. Set A and Set B

Step-by-step explanation:

We have been given three sets:

Set A: (5, 2), (4, 3), (3, 4), (2, 5)

Set B: (-1, -6), (0, 2), (1, 2), (3, 6)

Set C: (2, 1), (4, 2), (2, 3), (8, 4)

Now we need to state about which of the following sets of ordered pairs represent a function.

We know that a function can't have repeated values in domain that is x-value can't repeat.

we see that set C has repeated x-value "2".

Then set C is not a function.

Hence correct choice is:

d. Set A and Set B

vagabundo [1.1K]3 years ago

8 0

Answer:

A and B represent functions.

Step-by-step explanation:

In a function, any input (x-) value may have only one associated y value. If the same input appears more than once, we know immediately that the data do not represent a function.

A: The inputs are unique: {5, 4, 3, 2} so this is a function.

B: The inputs are unique: {-1, 0, 1, 3}, so this is a function.

C: 2 is used twice as input, so this is not a function.

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3 years ago

A player of a video game is confronted with a series of four opponents and an 80% probability of defeating each opponent. Assume

mixer [17]

Answer:

(a) 0.4096

(b) 0.64

Step-by-step explanation:

The probability that the player wins is,

$P(W)=0.80$

Then the probability that the player losses is,

$P(L)=1-P(W)=1-0.80=0.20$

The player is playing the video game with 4 different opponents.

It is provided that when the player is defeated by an opponent the game ends.

All the possible ways the player can win is: {L, WL, WWL, WWWL and WWWW)

(a)

The results from all the 4 opponents are independent, i.e. the result of a game played with one opponent is unaffected by the result of the game played with another opponent.

The probability that the player defeats all four opponents in a game is,

P (Player defeats all 4 opponents) = $P(W)\times P(W)\times P(W)\times P(W)=[P(W)]^{4} =(0.80)^{4}=0.4096$

Thus, the probability that the player defeats all four opponents in a game is 0.4096.

(b)

The probability that the player defeats at least two opponents in a game is,

P (Player defeats at least 2) = 1 - P (Player losses the 1st game) - P (Player losses the 2nd game) = $1-P(L)-P(WL)$

$=1-(0.20)-(0.80\times0.20)\\=1-0.20-0.16\\=0.64$

Thus, the probability that the player defeats at least two opponents in a game is 0.64.

(c)

Let X = number of times the player defeats all 4 opponents.

The probability that the player defeats all four opponents in a game is,

P(WWWW) = 0.4096.

Then the random variable $X\sim Bin(n=3, p=0.4096)$

The probability distribution of binomial is:

$P(X=x)={n\choose x}p^{x} (1-p)^{n-x}$

The probability that the player defeats all the 4 opponents at least once is,

P (X ≥ 1) = 1 - P (X < 1)

= 1 - P (X = 0)

$=1-[{3\choose 0}(0.4096)^{0} (1-0.4096)^{3-0}]\\=1-[1\times1\times (0.5904)^{3}\\=1-0.2058\\=0.7942$

Thus, the probability that the player defeats all the 4 opponents at least once is 0.7942.

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