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3 years ago

Choose the equation that represents the graph below: Graph of a line passing through points 0 comma 6 and 9 comma 0

Mathematics

1 answer:

kipiarov [429]3 years ago

6 0

By what you have described, the equation you're looking for would be

y = -(2/3)x+ 6

Rearrange if necessary.

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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

Is the pie of 34 rational are in rational

nexus9112 [7]

Answer:

34 would be a Rational number

Step-by-step explanation:

it's clean number and not sloppy like 0.888883

4 0

3 years ago

if 5 billion pennies are minted each year, how long would it take to mint the pennies needed to stack to the sun? plsss show all

dimulka [17.4K]

98 years.
the sun is 432,690 away and a penny is 1.52 mm in thickness
and 1.52 x 5 billion is 7,600,000,000 mm which is 4722.4 miles a year

6 0

2 years ago

Four less than 3 times a number at most is five

Jet001 [13]

The correct answer is the first one

8 0

3 years ago

Calculate the volume of the pyramid shown with measurements a = 3, b = 7, and c = 9.63 units348 units331.5 units330 units3

Yuri [45]

The volume of a pyramid is given by:

$V=\frac{1}{3}\cdot A_b\cdot h\text{.}$

Where:

• Ab is the area of the base,

• h is the height.

In this case, we have:

• a = 3 units,

• b = 7 units,

• c = 9 units,

• Ab = b * c,

• h = a = 3 units.

Replacing the data in the formula above, we get:

$V=\frac{1}{3}\cdot(b\cdot c)\cdot a=\frac{1}{3}\cdot(7\text{ units}\cdot9\text{ units})\cdot3\text{ units }=63units^3\text{.}$

Answer: 63 units³

5 0

1 year ago