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AlexFokin [52]
2 years ago
5

You randomly choose one of the morales.Without replacing the first marble, you choose a second marble. What is the probability o

f choosing red and then green?
P(Red and green)=1/4*1/4=1/16

(I really need help) ​

Mathematics
1 answer:
IrinaK [193]2 years ago
4 0
Let’s say, hypothetically speaking, you chose the second marble without replacing the first marble so, events are hypothetically dependent. Events are dependent if the occurrence of one hypothetical event hypothetically does affect the likelihood that the other events occur. The probably of two or more dependent events A and B is the probability of A times the probability of B after A hypothetically occurs

P(A and B) = P(A) x P(B after A)

Choose the first marble

The total number of hypothetical marbles are, hypothetically speaking, 4 on a hypothetical basis, and there is one red marble.

P(red)=1/4

Choose the second marble

Without hypothetically replacing the hypothetical first marble, you choose the hypothetical marble, hypothetically speaking. So, the total hypothetical number of marbles are, hypothetically, 3, and there is, hypothetically, one green marble.

P(green) = 1/3

The probability of choosing red and then, hypothetically, green is:

P(red and green) = P(red) x P(green)
=1/4 x 1/3

= 1/12

P(red and green) is hypothetically equal to 1/12 on a hypothetical account.

Final hypothetical answer: 1/12
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Answer:

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Step-by-step explanation:

Without Mincing words let us dive straight into the solution to the question above. We are given the following information which is going to aid in solving this particular question.

====> It is given, that there are 5 white balls and 10 red balls. Hence, the number of the total balls = 5 white balls + 10 red balls = 15 balls.

Therefore, the probability that 5 randomly selected balls contain exactly 3 white balls = \left[\begin{array}{ccc}5\\3\end{array}\right]  × \left[\begin{array}{ccc}10\\2&\end{array}\right]  ÷  \left[\begin{array}{ccc}15\\3\\\end{array}\right]  = 450 ÷ 3003 = 0.15

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3 years ago
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Answer:

\boxed{y=2x^2+4x-7}

Step-by-step explanation:

Let the quadratic function be

y=ax^2+bx+c


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We substitute (0,-7) to obtain;


-7=a(0)^2+b(0)^2+c


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We finally substitute (1,-1) to obtain;


-1=a(1)^2+b(1)^2+c


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We put equation (2) into equation (1) to get;


9=16a-4b-7


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\Rightarrow -1=a+b-7


\Rightarrow a+b=6---(5)


We add equation (4) and (5) to get;

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We put a=2 into equation (5) to get;


2+b=6


\Rightarrow b=6-2


\Rightarrow b=4


The reqiured quadratic function is

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