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Pavel [41]
3 years ago
14

There are 8 action, 3 comedy, and 5 drama DVDs on a shelf. Suppose three DVDs are selected at random from the shelf. Find each p

robability.
P(2 action, then a comedy), without replacement.
Mathematics
1 answer:
nordsb [41]3 years ago
4 0
8 action, 3 comedy, 5 drama....total = 16 dvd's

P(2 action, then a comedy)...without replacement

8/16 * 7/15 * 3/14 = (8 * 7 * 3) / (16 * 15 * 14) = 168/3360 = 1/20
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how many 5-digit even numbers can be formed with the digits 1,2,3,4,5, and 6 if repetition is allowed?​
Setler79 [48]

Given:

The given digits are 1,2,3,4,5, and 6.

To find:

The number of 5-digit even numbers that can be formed by using the given digits (if repetition is allowed).

Solution:

To form an even number, we need multiples of 2 at ones place.

In the given digits 2,4,6 are even number. So, the possible ways for the ones place is 3.

We have six given digits and repetition is allowed. So, the number of possible ways for each of the remaining four places is 6.

Total number of ways to form a 5 digit even number is:

Total=6\times 6\times 6\times 6\times 3

Total=3888

Therefore, total 3888 five-digit even numbers can be formed by using the given digits if repetition is allowed.

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2 years ago
Find the value of y and find m∠C and m∠D in the triangle given below.
stepan [7]

Answer:

Step-by-step explanation:

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2 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.

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A jewelry store is selling a set of 4 pairs of gemstone earrings for $58, including tax. Neva and three of her friends want to b
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