Factor the expression using the GCF. 44−11

You play in a soccer tournament, that consists of 5 games. Each game you win with probability .6, lose with probability .3, and

nasty-shy [4]

Answer:

(a) The joint PMF of W, L and T is:

$P(W,\ L,\ T)={5\choose (n_{W}!\times n_{L}!\times n_{T}!)}\times [0.60]^{n_{W}}\times [0.30]^{n_{L}}\times [0.10]^{n_{T}}$

(b) The marginal PMF of W is:

$P(W=w)={5\choose n_{W}!}\times 0.60^{n_{W}}\times (1-0.60)^{n-n_{W}}$

Step-by-step explanation:

Let X = number of soccer games played.

The outcome of the random variable X are:

W = if a game won

L = if a game is lost

T = if there is a tie

The probability of winning a game is, P (W) = 0.60.

The probability of losing a game is, P (L) = 0.30.

The probability of a tie is, P (T) = 0.10.

The sum of the probabilities of the outcomes of X are:

P (W) + P (L) + P (T) = 0.60 + 0.30 + 0.10 = 1.00

Thus, the distribution of W, L and T is a appropriate probability distribution.

(a)

Now, the outcomes W, L and T are one experiment.

The distribution of n independent and repeated trials, each having a discrete number of outcomes, each outcome occurring with a distinct constant probability is known as a Multinomial distribution.

The outcomes of X follows a Multinomial distribution.

The joint probability mass function of W, L and T is:

$P(W,\ L,\ T)={n\choose (n_{W}!\times n_{L}!\times n_{T}!)}\times [P(W)]^{n_{W}}\times [P(L)]^{n_{L}}\times [P(T)]^{n_{T}}$

The soccer tournament consists of n = 5 games.

Then the joint PMF of W, L and T is:

$P(W,\ L,\ T)={5\choose (n_{W}!\times n_{L}!\times n_{T}!)}\times [0.60]^{n_{W}}\times [0.30]^{n_{L}}\times [0.10]^{n_{T}}$

(b)

The random variable W is defined as the number games won in the soccer tournament.

The probability of winning a game is, P (W) = p = 0.60.

Total number of games in the tournament is, n = 5.

A game is won independently of the others.

The random variable W follows a Binomial distribution.

The probability mass function of W is:

$P(W=w)={5\choose n_{W}!}\times 0.60^{n_{W}}\times (1-0.60)^{n-n_{W}}$

Thus, the marginal PMF of W is:

$P(W=w)={5\choose n_{W}!}\times 0.60^{n_{W}}\times (1-0.60)^{n-n_{W}}$

3 0

3 years ago

3.Tracie set out from Elmhurst, IL to go to Franklin Park. On the way, she made a few stops to do errands. Each stop is indicate

fenix001 [56]

The total distance Tracy traveled was 15,000 feet assuming each unit represents 1,000 feet.

<h3>How to calculate the total distance?</h3>

The total distance is the sum of all the distance between stops.

Partial distances:

Home to stop 1: 2,000 feet
Stop 1 to 2: 4,000 feet
Stop 2 to 3: 5,000 feet
Stop 3 to 4: 4,000 feet
Total distance: 15,000 feet

However, the real distance would be 10.63 units based on the locations in the grid.

d=√(0−8)^2+(7−0)^2
d=√8^2+7^2
d=√64+49
d=√113 =10.63 units

Learn more about feets in: brainly.com/question/15838370

3 0

2 years ago

There are 4 different methods you could use to solve a quadratic equation that would depending upon the actual equation. They ar

lana66690 [7]

Answer:

Part A: x^2 + 6x + 8 = 0 use the factoring the equation method

Part A: x^2 + 6x + 8 = 0 use the quadratic formula

Step-by-step explanation:

Part A;

The equation is x^2 + 6x + 8 = 0 , looking at this quadratic expression, you notice it is written in a quadratic equation standard form of ax^2+bx+c=0. Additionally, you notice that can find what multiplied to get the quadratic equation,factor.You can identify two numbers that multiply to get ac and add to give b.In this question;

a=1,b=6,c=8

ac=8

The numbers are 4 and 2. Factoring the equation method will give;

x²+6x+8=0

x²+4x+2x+8=0

x(x+4)+2(x+4)=0

(x+2)+(x+4)=x²+6x+8

x+2=0, x=-2 and x+4=0, x=-4

Part B

The quadratic equation is ;

x²+6x-11=0

You notice that there are no factors that multiply direct to get the quadratic equation like in part 1. When you observe, a=1, b=6 and c=-11

ac=1×-11=-11 and b=6 .You notice there are no factors that multiply to give -11 and add to get 6, hence the factorizing the equation method can not be used.However, you can apply the quadratic formula that requires coefficients. You have a=1, b=c and c=-11 as the coefficients to use in the quadratic formula.

8 0

3 years ago

Read 2 more answers

Write a system of linear equations that has the ordered pair as its solution (3,5)

ololo11 [35]

An easy way to do this is by using the simplest equations you can:

x + y and x - y

(3, 5) is (x, y). all you have to do is plug those into your equations and get a result, so:

x + y = 8
(because 3 + 5 = 8)
x - y = -2
(because 3 - 5 = -2)

and those can serve as your system of equations.

you can check it by solving by substitution:
x - y = -2
x + y = 8

solve one of the equations for a single variable:
x = y - 2

plug it into the second equation:
(y - 2) + y = 8
y - 2 + y = 8
2y - 2 = 8
2y = 10
y = 5

then plug that result back into the equation:
x + y = 8
x + 5 = 8
x = 3

8 0

3 years ago

I am thinking of a number.

Step2247 [10]

Step-by-step explanation:

Let the number Noah is thinking of be x,

=> x - 19 * 4 = 16

=> x - 19 = 16/4

=> x = 16/4 + 19

=> x = 4 + 19

=> x = 23

If my answer helped , please mark me as brainliest,

Thank you

4 0

4 years ago