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Nutka1998 [239]
3 years ago
11

June father is six times as old as June the sum of their age is no less than 77 what is the youngest age June can be

Mathematics
1 answer:
alexgriva [62]3 years ago
6 0
Let June's age be
x \: years

Then his father is
6x \: years

The sum of their ages is no less than 77 means,
x + 6x \geqslant 77
7x \geqslant 77
x \geqslant 11
The youngest age June can be is 11
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Calculate the expected value, the variance, and the standard deviation of the given random variable X. (Round all answers to two
Tom [10]

Answer:

E (X) = 1.8

Var (X) = 0.36

σ = 0.6

Step-by-step explanation:

Solution:-

- Denote the random variable X : is the number of red marbles that Suzan has in her hand after she selects three marbles.

- Total sample space (bag) have the following quantity of colored marbles:

                Bag : { 3 Red , 2 Green }

- Suzan selects three marbles from the bag. The Event (X) defines the number of red marbles out of 3.

- The total number of outcomes / selections for randomly selecting 3 balls from the bag:

               All outcomes = 5 C 3 = 10

- The probability distribution of the random variable X, we will use combinations to determine the required probabilities:

  X = 1 red marble:

        P ( X = 1 ) : Suzan chooses 1 Red marble from the available 3 red marble and 2 green marbles.

        P ( X = 1 ) = [ 3C1*2C2 ] / all outcomes = (3*1) / 10 = 0.3

X = 2 red marble:

        P ( X = 2 ) : Suzan chooses 2 Red marble from the available 3 red marble and 1 green marbles.

        P ( X = 2 ) = [ 3C2*2C1 ] / all outcomes = (3*2) / 10 = 0.6

X = 3 red marble:

        P ( X = 3 ) : Suzan chooses 3 Red marble from the available 3 red marble.

        P ( X = 3 ) = [ 3C3] / all outcomes = (1) / 10 = 0.1

- The probability distribution is as follows:

          X :         1                 2                3

       P (X):       0.3            0.6              0.1

- The expected value E(X) for the given random variable X is:

                E ( X ) = ∑Xi*P(Xi)

                           = 1*0.3 + 2*0.6 + 3*0.1

                           =1.8

- The variance Var(X) for the given random variable X is:

                Var ( X ) = ∑Xi^2*P(Xi) - [ E(X) ] ^2

                               = 1^2*0.3 + 2^2*0.6 + 3^2*0.1 - 1.8^2

                               = 0.36

- The standard deviation for the given random variable X is:

                σ = √Var(X)

                σ = √0.36

                σ = 0.6

5 0
3 years ago
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