Whats the answer for -2(3n+6)=1+7n

A player of a video game is confronted with a series of opponents and has an 80% probability of defeating each one. Success with

Vikentia [17]

Answer:

(a) The PMF of X is: $P(X=k)=(1-0.20)^{k-1}0.20;\ k=0, 1, 2, 3....$

(b) The probability that a player defeats at least two opponents in a game is 0.64.

(c) The expected number of opponents contested in a game is 5.

(d) The probability that a player contests four or more opponents in a game is 0.512.

(e) The expected number of game plays until a player contests four or more opponents is 2.

Step-by-step explanation:

Let X = number of games played.

It is provided that the player continues to contest opponents until defeated.

(a)

The random variable X follows a Geometric distribution.

The probability mass function of X is:

$P(X=k)=(1-p)^{k-1}p;\ p>0, k=0, 1, 2, 3....$

It is provided that the player has a probability of 0.80 to defeat each opponent. This implies that there is 0.20 probability that the player will be defeated by each opponent.

Then the PMF of X is:

$P(X=k)=(1-0.20)^{k-1}0.20;\ k=0, 1, 2, 3....$

(b)

Compute the probability that a player defeats at least two opponents in a game as follows:

P (X ≥ 2) = 1 - P (X ≤ 2)

= 1 - P (X = 1) - P (X = 2)

$=1-(1-0.20)^{1-1}0.20-(1-0.20)^{2-1}0.20\\=1-0.20-0.16\\=0.64$

Thus, the probability that a player defeats at least two opponents in a game is 0.64.

(c)

The expected value of a Geometric distribution is given by,

$E(X)=\frac{1}{p}$

Compute the expected number of opponents contested in a game as follows:

$E(X)=\frac{1}{p}=\frac{1}{0.20}=5$

Thus, the expected number of opponents contested in a game is 5.

(d)

Compute the probability that a player contests four or more opponents in a game as follows:

P (X ≥ 4) = 1 - P (X ≤ 3)

= 1 - P (X = 1) - P (X = 2) - P (X = 3)

$=1-(1-0.20)^{1-1}0.20-(1-0.20)^{2-1}0.20-(1-0.20)^{3-1}0.20\\=1-0.20-0.16-0.128\\=0.512$

Thus, the probability that a player contests four or more opponents in a game is 0.512.

(e)

Compute the expected number of game plays until a player contests four or more opponents as follows:

$E(X\geq 4)=\frac{1}{P(X\geq 4)}=\frac{1}{0.512}=1.953125\approx 2$

Thus, the expected number of game plays until a player contests four or more opponents is 2.

4 0

3 years ago

B

Gnesinka [82]

Answer:

13

Step-by-step explanation:

We know that a^2+b^2 = c^2 where a and b are the legs and c is the hypotenuse

5^2 + 12^2 = c^2

25+144 = c^2

169 = c^2

Taking the square root of each side

sqrt(169) = sqrt(c^2)

13 =c

3 0

3 years ago

What pattern would to you expect when two variables are related by a linear model with a positive slope

olganol [36]

There isnt one its has to be consistent but honestly idk

4 0

3 years ago

In ∆ABC, ∠A is a right angle and m∠B = 45°. Find BC.

Allisa [31]

Step-by-step explanation:

In ∆ABC, if <A is 90° and <B is 45° then hypotenuse is BC.

Now,

Cos B= base/height

Cos B= AB/ BC

BC= AB/cos 45°.

4 0

4 years ago

A bank has a special vault for valuable items . It has 3 dials that operate the combination . The first dial has the numbers fro

SCORPION-xisa [38]

Answer:

There are 67,600 possible combinations for this vault.

Step-by-step explanation:

First dial: 100 possible values

Second dial: 26 possible values

Third dial: 26 possible values.

How many possible combinations are there for this vault?

Multiplication of these values. So

100*26*26 = 67600

There are 67,600 possible combinations for this vault.

7 0

3 years ago