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Iteru [2.4K]
3 years ago
11

Identify the mapping diagram that represents the relation and determine whether the relation is a function. {(-8, -6), (-5, 2),

(-8, 1), (7, 3)} A. The relation is a function. B. The relation is not a function C. The relation is a function. D. The relation is not a function.

Mathematics
2 answers:
Deffense [45]3 years ago
8 0

Answer:

Option D is correct.

The relation {(-8, -6) (-5, 2) (-8, 1) (7, 3)} is not a function.

Step-by-step explanation:

Given the relation:  {(-8, -6) (-5, 2) (-8, 1) (7, 3)}

Domain is the set of all possible inputs of a relation i.e { -8, -5 , -8 , 7}

Range is the set of output values of a function i.e, {-6, 2 , 1 , 3}

The mapping as shown below in the figure:

A function is a relation in which every element of the domain is matched to not more than one element of the range.

In other words, we can say that ,no value of x gets mapped to more than 1 value of y.

Since, from the mapping you can see that the domain value -8  paired with -6 and 1;  as x is used more than once.

Therefore, this relation is not a function

My name is Ann [436]3 years ago
4 0
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Delvig [45]

Answer:

a) 182 possible ways.

b) 5148 possible ways.

c) 1378 possible ways.

d) 2899 possible ways.

Step-by-step explanation:

The order in which the cards are chosen is not important, which means that we use the combinations formula to solve this question.

Combinations formula:

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)!}

In this question, we have that:

There are 52 total cards, of which:

13 are spades.

13 are diamonds.

13 are hearts.

13 are clubs.

(a)Two-pairs: Two pairs plus another card of a different value, for example:

2 pairs of 2 from sets os 13.

1 other card, from a set of 26(whichever two cards were not chosen above). So

T = 2C_{13,2} + C_{26,1} = 2*\frac{13!}{2!11!} + \frac{26!}{1!25!} = 182

So 182 possible ways.

(b)Flush: five cards of the same suit but different values, for example:

4 combinations of 5 from a set of 13(can be all spades, all diamonds, and hearts or all clubs). So

T = 4*C_{13,5} = 4*\frac{13!}{5!8!} = 5148

So 5148 possible ways.

(c)Full house: A three of a kind and a pair, for example:

4 combinations of 3 from a set of 13(three of a kind ,c an be all possible kinds).

3 combinations of 2 from a set of 13(the pair, cant be the kind chosen for the trio, so 3 combinations). So

T = 4*C_{13,3} + 3*C_{13.2} = 4*\frac{13!}{3!10!} + 3*\frac{13!}{2!11!} = 1378

So 1378 possible ways.

(d)Four of a kind: Four cards of the same value, for example:

4 combinations of 4 from a set of 13(four of a kind, can be all spades, all diamonds, and hearts or all clubs).

1 from the remaining 39(do not involve the kind chosen above). So

T = 4*C_{13,4} + C_{39,1} = 4*\frac{13!}{4!9!} + \frac{39!}{1!38!} = 2899

So 2899 possible ways.

4 0
3 years ago
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