MATHEMATICS HIGH SCHOOL

Determine whether each relation represents a function. For each function, state the domain and range. 1) {(2,6), (-3,6), (4,9), (1,10)}
2) {(1,3), (2,3), (3,3), (4,3)}
3) {(-2,4), (-2,6), (0,3), (3,7)}
4) {(-2,4), (-1,1), (0,0), (1,1)}

Answers

Answer 1

Answer:

Option 1 , 2 and 4 is defined as a function.

And, for 1 relation;

The domain of relation = {2, -3, 4, 1}

And, Range of relation = {6, 9 ,10}

For 2 relation;

The domain of relation= {1, 2, 3, 4}

The range of relation = {3}

For 4 relation ;

The domain of the relation = {-2, -1, 0, 1}

The range of the relation = {4, 1, 0}

What is function?

A function is defined as a relation between a set of input having only one output.

Now,

The relation is;

1) {(2,6), (-3,6), (4,9), (1,10)}

Hence, The domain of relation = {2, -3, 4, 1}

And, Range of relation = {6, 9 ,10}

Clearly, Each input (domain) has only one output (range)

Hence, The given relation is function.

For,The relation;

2) {(1,3), (2,3), (3,3), (4,3)}

The domain of relation= {1, 2, 3, 4}

The range of relation = {3}

Clearly, Each input (domain) has only one output (range).

Hence, The given relation is function.

For, The function;

3) {(-2,4), (-2,6), (0,3), (3,7)}

Clearly, -2 has two image 4 and 6.

So, This relation is not a function.

For, The function;

4) {(-2,4), (-1,1), (0,0), (1,1)}

The domain of the relation = {-2, -1, 0, 1}

The range of the relation = {4, 1, 0}

Clearly, Each input (domain) has only one output (range).

Hence, This relation is a function.

Therefore,

Option 1 , 2 and 4 is defined as a function.

And, for 1 relation;

The domain of relation = {2, -3, 4, 1}

And, Range of relation = {6, 9 ,10}

For 2 relation;

The domain of relation= {1, 2, 3, 4}

The range of relation = {3}

For 4 relation ;

The domain of the relation = {-2, -1, 0, 1}

The range of the relation = {4, 1, 0}

Learn more about the function visit:

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Answer 2

Answer: $D-domain;\ R-range\n\n1)\ \{(2;\ 6),\ (-3;\ 6),\ (4;\ 9),\ (1;\ 10)\}\ YES\ :)\nD=\{2;-3;\ 4;\ 1\};\ R=\{6;\ 9;\ 10\}\n\n2)\ \{(1;\ 3),\ (2;\ 3),\ (3;\ 3),\ (4;\ 3)\}\ \ \ YES\ :)\nD=\{1;\ 2;\ 3;\ 4\};\ R=\{3\}\n\n3)\ \{(\fbox{-2};\ 4),\ (\fbox{-2};\ 6),\ (0;\ 3),\ (3;\ 7)\}\ \ \ NO\ :(\n\n4)\ \{(-2;\ 4),\ (-1;\ 1),\ (0;\ 0),\ (1;\ 1)\}\ \ \ \ YES\ :)\nD=\{-2;-1;\ 0;\ 1\};\ R=\{4;\ 1;\ 0\}$

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$10:(3)/(5)=10 \cdot (5)/(3)=(50)/(3)=16(2)/(3)\n \nAnswer : \ 16 \ complete \ bowls$

$(3)/(5) = 0.6$

$(0.6)/(1) = (10)/(x)$

$0.6x=10$

$x= (10)/(0.6)$

$x=16 (2)/(3)$

16 complete bowls or 16 2/3 bowls

at a high school, the probability that a student is a senior is 0.25. the probability that a student plays a sport is 0.20. the probability that a student is a senior and plays a sport is 0.08. what is the probability that a randomly selected student plays a sport, given that the student is a senior?

Answers

Answer:

0.32

Step-by-step explanation:

We have been given that at a high school, the probability that a student is a senior is 0.25. The probability that a student plays a sport is 0.20. The probability that a student is a senior and plays a sport is 0.08.

We will use conditional probability formula to solve our given problem. $P(B|A)=\frac{P(\text{A and B)}}{P(A)}$ , where,

$P(B|A)$ = The probability of event B given event A.

$P(\text{A and B)}$ = The probability of event A and event B.

$P(A)$ =Probability of event A.

Let A be that the student is senior and B be the student plays a sport.

P(A and B) = Probability that student is a senior and plays a sport.

$P(B|A)=\frac{\text{Probability that a student is a senior and plays a sport}}{\text{Probability that a student is senior}}$

Upon substituting our given values we will get,

$P(B|A)=(0.08)/(0.25)$

$P(B|A)=0.32$

Therefore, the probability that a randomly selected student plays a sport, given that the student is a senior will be 0.32.

If xy < zy < 0, is y positive? x < z x is negativeStatement (1) ALONE is sufficient, but statement (2) alone is not sufficient to answer the question asked
Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient to answer the question asked
Both statements (1) and (2) TOGETHER are sufficient to answer the question asked; but NEITHER statement ALONE is sufficient

Answers

Answer:

Both statements (1) and (2) TOGETHER are sufficient to answer the question asked

Step-by-step explanation:

Given statement 1 :

xy < zy < 0,

The product of two numbers are negative if either of the numbers are negative.

∵ if xy < 0 ⇒ Case 1 : x > 0 and y < 0

Case 2 : x < 0 and y > 0,

Thus, Statement is not sufficient to prove y is positive,

Now, Statement 2 :

x < z, x is negative,

That is, x < 0

Combining statements (1) and (2),

We get,

xy < 0, x < 0,

⇒ y > 0

That is, y is positive.

Hence, Both statements (1) and (2) TOGETHER are sufficient to answer the question asked

Which is a point 6units to the left of 4

Answers

-2 i believe, have a great day :)

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Answers

Answer: -1/3

Step-by-step explanation:

gradient = change in y/change in x

gradient = 7-8/6-3

gradient = -1/3

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Remember order of operations. Please (Parenthesis and brackets) Excuse (exponents) My (multiplication) Dear (Division) Aunt (Add) Sally (Subtract). Also remember that multiplication and division as well as addition and subtraction should go from left to right. So if you were to do this problem first I'd convert the fraction 1/10 to a decimal. That's like saying 1 in 10 or 10%. So it should convert to .10 x 6 = 0.60 and don't forget to multiple 7 x1000 next. (Left to right). 7,000. So it becomes 5+10+0.60+7,000 then add. 7,015.60 is your answer.

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Determine whether each relation represents a function. For each function, state the domain and range. 1) {(2,6), (-3,6), (4,9), (1,10)} 2) {(1,3), (2,3), (3,3), (4,3)} 3) {(-2,4), (-2,6), (0,3), (3,7)} 4) {(-2,4), (-1,1), (0,0), (1,1)}

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Determine whether each relation represents a function. For each function, state the domain and range. 1) {(2,6), (-3,6), (4,9), (1,10)}
2) {(1,3), (2,3), (3,3), (4,3)}
3) {(-2,4), (-2,6), (0,3), (3,7)}
4) {(-2,4), (-1,1), (0,0), (1,1)}