For a function p(x), as — then p(x) — 4. Which of the following graphs could be p(x)?

The kindergarten class has 10 girls and 10 boys. If two students are chosen at random to be in the school play, what is the prob

Akimi4 [234]

50% because there's equal value of both variables

6 0

3 years ago

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Which Trigonometric functions have a domain of (-1,1)

Ilia_Sergeevich [38]

Answer:

Step-by-step explanation:

[−1,1] and [0, $\pi$ ]

hope this helped :)

8 0

3 years ago

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Equation for ( 6, 13) ( 7900, 5 )

tia_tia [17]

For this case we have that by definition, the equation of the line in the slope-intersection form is given by:

$y=mx+b$

Where:

m: It is the slope of the line

b: It is the cut-off point with the y axis

We have the following points through which the line passes:

$(x_ {1}, y_ {1}) :( 6,13)\\(x_ {2}, y_ {2}): (7900,5)$

We find the slope of the line:

$m = \frac {y_ {2} -y_ {1}} {x_ {2} -x_ {1}} = \frac {5-13} {7900-6} = \frac {-8} {7894} = - \frac {4} {3947}$

Thus, the equation of the line is of the form:

$y = - \frac {4} {3947} x + b$

We substitute one of the points and find b:

$13 = - \frac {4} {3947} (6) + b\\13 = - \frac {24} {3947} + b\\13+ \frac {24} {3947} = b\\b = \frac {51335} {3947}$

Finally, the equation is:

$y = - \frac {4} {3947} x + \frac {51335} {3947}$

Answer:

$y = - \frac {4} {3947} x + \frac {51335} {3947}$

5 0

3 years ago

According to the Rational Root Theorem, which number is a potential root of f(x) = 9x

MAVERICK [17]

Answer:

The answer is below....V

Step-by-step explanation:

Brainliest when you can if you will!!!!

5 0

3 years ago

Given a standard deck of 52 cards, 3 cards are dealt without replacement. Using this situation, answer the questions below.<b

kherson [118]

Given that 3 cards are dealt without replacement in a standard deck of 52 cards.

Part A:

There are 4 queens in a standard deck of 52 card, thus the probability that the first card is a queen is given by 4 / 52 = 1 / 13.

Since, the first card is not replaced, thus there are 3 queens remaining and 51 ards remaining in total, thus the probability that the second card is a queen is given by 3 / 51 = 1 / 17

Similarly the probability that the third card is a queen is given by 2 / 50 = 1 / 25.

Therefore, the probability that all three cards are queens is given by

$\frac{1}{13} \times \frac{1}{17} \times \frac{1}{25} = \frac{1}{5525}$

Part B:

Yes the probability of drawing a queen of heart is independent of the probability of drawing a queen of diamonds because they are separate cards and drawing one of the cards does not in any way affect the chance of drawing the other card.

Part C:

Given that the first card is a queen, then there are 3 queens remaining out of 51 cards remaining, thus the number of cards that are not queen is 51 - 3 = 48 cards.

Therefore, if the first card is a queen, the probability that the second card will not be a queen is given by 48 / 51 = 16 / 17

Part D:

Given that the first two card are queens, then there are 2 queens remaining out of 50 cards remaining.

Therefore, if two of the three cards are queens ,the probability that you will be dealt three queens is given by 2 / 50 = 1 / 25 = 0.04

Part E:

Given that the first two card are queens, then there are 2 queens remaining out of 50 cards remaining, thus the number of cards that are not queen is 50 - 2 = 48 cards.

Therefore, if two of the three cards are queens ,the probability that the other card is not a queen is given by 48 / 50 = 24 / 25 = 0.96

8 0

3 years ago