Is k6= a solution or non solution

Find the output, k, when the input, T. iS<br> 5.<br> 67 + 100

rjkz [21]

Answer:

70

Step-by-step explanation:

k = 6(-5)+100 = -30+100 = 70

3 0

2 years ago

A random variableX= {0, 1, 2, 3, ...} has cumulative distribution function.a) Calculate the probability that 3 ≤X≤ 5.b) Find the

olchik [2.2K]

Answer:

a) P ( 3 ≤X≤ 5 ) = 0.02619

b) E(X) = 1

Step-by-step explanation:

Given:

- The CDF of a random variable X = { 0 , 1 , 2 , 3 , .... } is given as:

$F(X) = P ( X =< x) = 1 - \frac{1}{(x+1)*(x+2)}$

Find:

a.Calculate the probability that 3 ≤X≤ 5

b) Find the expected value of X, E(X), using the fact that. (Hint: You will have to evaluate an infinite sum, but that will be easy to do if you notice that

Solution:

- The CDF gives the probability of (X < x) for any value of x. So to compute the P ( 3 ≤X≤ 5 ) we will set the limits.

$F(X) = P ( 3=$

- The Expected Value can be determined by sum to infinity of CDF:

E(X) = Σ ( 1 - F(X) )

$E(X) = \frac{1}{(x+1)*(x+2)} = \frac{1}{(x+1)} - \frac{1}{(x+2)} \\\\= \frac{1}{(1)} - \frac{1}{(2)}\\\\= \frac{1}{(2)} - \frac{1}{(3)} \\\\= \frac{1}{(3)} - \frac{1}{(4)}\\\\= ............................................\\\\= \frac{1}{(n)} - \frac{1}{(n+1)}\\\\= \frac{1}{(n+1)} - \frac{1}{(n+ 2)}$

E(X) = Limit n->∞ [1 - 1 / ( n + 2 ) ]

E(X) = 1

8 0

3 years ago

25-28 i need help with

valina [46]

Don't understand the question your asking

3 0

3 years ago

A theater has a seating capacity of 750 and charges $3 for children, s5 for students, and $7 for adults. At a certain screening

-BARSIC- [3]

Answer:

Between 150 and 450

Step-by-step explanation:

We are going to find the number by resolving a system of linear equations.

First we write the system equations :

$C+S+A=750$

Where C : children, S : students and A : adults

The equation represents the ''full attendance''

The second equation :

$3C+5S+7A=3450$

This equation represents the totaled receipts.

The system :

$C+S+A=750\\3C+5S+7A=3450$

has the following associated matrix :

$\left[\begin{array}{cccc}1&1&1&750\\3&5&7&3450\end{array}\right]$

By performing elementary matrix operations we find that the matrix is equivalent to

$\left[\begin{array}{cccc}1&0&-1&150\\0&1&2&600\\\end{array}\right]$

The new system :

$C-A=150\\S+2A=600$

Working with the equations :

$C = 150 + A\\S = 600-2A$

Our solution vector is :

$\left[\begin{array}{c}C&S&A\end{array}\right] =\left[\begin{array}{c}150+A&600-2A&A\end{array}\right]$

For example :

If 0 adults attended ⇒ A = 0

$C = 150 + 0 \\C = 150\\S = 600 - 2A\\S = 600$

This verify the totaled receipts equation :

150($3)+600($5) = $ 3450

A ≥ 0 ⇒ If A = 0 ⇒ C = 150

C = 150 is the minimum children attendance

From the equation :

$S = 600 -2A$

S ≥0

600 - 2A ≥ 0

600 ≥ 2A

300≥ A

A is restricted to the interval [ 0, 300]

When A = 0 ⇒ C = 150

When A = 300 ⇒C = 150 + A = 150 + 300 = 450

Children ∈ [ 150,450]

With C being an integer number (including 0)

Also S and A are integer numbers (including 0)

7 0

3 years ago

A blackjack player at a Las Vegas casino learned that the house will provide a free room if play is for four hours at an average

marysya [2.9K]

Answer:

a) player’s expected payoff is $ 240

b) probability the player loses $1000 or more is 0.1788

c) probability the player wins is 0.3557

d) probability of going broke is 0.0594

Step-by-step explanation:

Given:

Since there are 60 hands per hour and the player plays for four hours then the sample size is:

n = 60 * 4 = 240

The player’s strategy provides a probability of .49 of winning on any one hand so the probability of success is:

p = 0.49

a)

Solution:

Expected payoff is basically the expected mean

Since the bet is $50 so $50 is gained when the player wins a hand and $50 is lost when the player loses a hand. So

Expected loss = μ

= ∑ x P(x)

= 50 * P(win) - 50 * P(lose)

= 50 * P(win) + (-50) * (1 - P(win))

= 50 * 0.49 - 50 * (1 - 0.49)

= 24.5 - 50 ( 0.51 )

= 24.5 - 25.5

= -1

Since n=240 and expected loss is $1 per hand then the expected loss in four hours is:

240 * 1 = $ 240

b)

Using normal approximation of binomial distribution:

n = 240

p = 0.49

q = 1 - p = 1 - 0.49 = 0.51

np = 240 * 0.49 = 117.6

nq = 240 * 0.51 = 122.5

both np and nq are greater than 5 so the binomial distribution can be approximated by normal distribution

Compute z-score:

z = x - np / √(np(1-p))

= 110.5 - 117.6 / √117.6(1-0.49)

= −7.1/√117.6(0.51)

= −7.1/√59.976

= −7.1/7.744417

=−0.916789

Here the player loses 1000 or more when he loses at least 130 of 240 hands so the wins is 240-130 = 110

Using normal probability table:

P(X≤110) = P(X<110.5)

= P(Z<-0.916)

= 0.1788

c)

Using normal approximation of binomial distribution:

n = 240

p = 0.49

q = 1 - p = 1 - 0.49 = 0.51

np = 240 * 0.49 = 117.6

nq = 240 * 0.51 = 122.5

both np and nq are greater than 5 so the binomial distribution can be approximated by normal distribution

Compute z-score:

z = x - np / √(np(1-p))

= 120.5 - 117.6 / √117.6(1-0.49)

= 2.9/√117.6(0.51)

= 2.9/√59.976

= 2.9/7.744417

=0.374463

Here the player wins when he wins at least 120 of 240 hands

Using normal probability table:

P(X>120) = P(X>120.5)

= P(Z>0.3744)

= 1 - P(Z<0.3744)

= 1 - 0.6443

= 0.3557

d)

Player goes broke when he loses $1500

Using normal approximation of binomial distribution:

n = 240

p = 0.49

q = 1 - p = 1 - 0.49 = 0.51

np = 240 * 0.49 = 117.6

nq = 240 * 0.51 = 122.5

both np and nq are greater than 5 so the binomial distribution can be approximated by normal distribution

Compute z-score:

z = x - np / √(np(1-p))

= 105.5 - 117.6 / √117.6(1-0.49)

= -12.1/√117.6(0.51)

= -12.1/√59.976

= -12.1/7.744417

=−1.562416

Here the player loses 1500 or more when he loses at least 135 of 240 hands so the wins is 240-135 = 105

Using normal probability table:

P(X≤105) = P(X<105.5)

= P(Z<-1.562)

= 0.0594

7 0

3 years ago