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Sedaia [141]
3 years ago
12

A bag has four green marbles, three red marbles, and three yellow marbles. What is the probability that you pick a red marble, d

o not replace it, and pick a yellow marble?
2/15 1/15 1/9 1/10
Mathematics
2 answers:
Natasha2012 [34]3 years ago
7 0

Answer:

1/10

Step-by-step explanation:

four green marbles, three red marbles, and three yellow marbles

Original total number of marbles: 4 + 3 + 3 = 10

First pick

Picking a red marble. There are 3 red marbles.

p(red) = 3/10

Second pick

Since the red marble picked is not replaced, now the total is 9 marbles.

Picking a yellow marble. There are 3 yellow marbles.

p(yellow) = 3/9 = 1/3

The probability of both events is the product of the individual probabilities.

p(red then yellow) = 3/10 * 1/3 = 1/10

Archy [21]3 years ago
5 0
The probability is 1/9
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lakkis [162]

Answer:

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Step-by-step explanation:

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6 0
3 years ago
The diameter of a particle of contamination (in micrometers) is modeled with the probability density function f(x)= 2/x^3 for x
natulia [17]

Answer:

a) 0.96

b) 0.016

c) 0.018

d) 0.982

e) x = 2

Step-by-step explanation:

We are given with the Probability density function f(x)= 2/x^3 where x > 1.

<em>Firstly we will calculate the general probability that of P(a < X < b) </em>

       P(a < X < b) =  \int_{a}^{b} \frac{2}{x^{3}} dx = 2\int_{a}^{b} x^{-3} dx

                            = 2[ \frac{x^{-3+1} }{-3+1}]^{b}_a   dx    { Because \int_{a}^{b} x^{n} dx = [ \frac{x^{n+1} }{n+1}]^{b}_a }

                            = 2[ \frac{x^{-2} }{-2}]^{b}_a = \frac{2}{-2} [ x^{-2} ]^{b}_a

                            = -1 [ b^{-2} - a^{-2}  ] = \frac{1}{a^{2} } - \frac{1}{b^{2} }

a) Now P(X < 5) = P(1 < X < 5)  {because x > 1 }

     Comparing with general probability we get,

     P(1 < X < 5) = \frac{1}{1^{2} } - \frac{1}{5^{2} } = 1 - \frac{1}{25} = 0.96 .

b) P(X > 8) = P(8 < X < ∞) = 1/8^{2} - 1/∞ = 1/64 - 0 = 0.016

c) P(6 < X < 10) = \frac{1}{6^{2} } - \frac{1}{10^{2} } = \frac{1}{36} - \frac{1}{100 } = 0.018 .

d) P(x < 6 or X > 10) = P(1 < X < 6) + P(10 < X < ∞)

                                = (\frac{1}{1^{2} } - \frac{1}{6^{2} }) + (1/10^{2} - 1/∞) = 1 - 1/36 + 1/100 + 0 = 0.982

e) We have to find x such that P(X < x) = 0.75 ;

               ⇒  P(1 < X < x) = 0.75

               ⇒  \frac{1}{1^{2} } - \frac{1}{x^{2} } = 0.75

               ⇒  \frac{1} {x^{2} } = 1 - 0.75 = 0.25

               ⇒  x^{2} = \frac{1}{0.25}   ⇒ x^{2} = 4 ⇒ x = 2  

Therefore, value of x such that P(X < x) = 0.75 is 2.

8 0
2 years ago
30 POINTS TO WHOEVER HELPS WITH THIS I NEED IT ASAP PLEASE HELP
Mekhanik [1.2K]

Answer:

<em>4a ) Difference ⇒ $ 63.24,</em>

<em>4b ) 4 children</em>

Step-by-step explanation:

4a ) If the number of children is represented by the value of x, let us substitute a family of 2 children as value of x into this equation we are given;

y = 21.08 * ( 2 ) + 85.15 ⇒ Multiply 21.08 by 2,

y = 42.16 + 85.15 ⇒ Combine like terms,

y = 127.31

If y ⇒ amount spent on grocery's, let us solve for the amount spent on 5 children, and subtract from this amount;

y = 21.08 * ( 5 ) + 85.15 ⇒ Multiply 21.08 by 5,

y = 105.4 + 85.15 ⇒ Combine like terms,

y = 190.55

Difference; 190.55 - 127.31 = 63.24,

<em>Solution; Difference ⇒ $ 63.24</em>

4b ) If the amount is $ 169.47, let us plug it into our given equation provided amount spent on groceries ⇒ y;

169.47 = 21.08x + 85.15 ⇒ Subtract 85.15 from either side of equation,

21.08x = 84.32 ⇒ Divide either side 21.08,

x = 4 children

<em>Solution; 4 children</em>

5 0
3 years ago
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andreev551 [17]

9) They traveled 31.9 hours over the 3-day trip.

10.9 + 8.6 + 12.4 = 31.9

10) She does not have enough money. She has $12, but everything she wants to purchase would cost $12.39.

Large drink = 1.79

Turkey sandwich = 4.85

1.79 + 4.85 = 6.64

6.64 + 5.75 = 12.39

5 0
3 years ago
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