Simplify the expression: Look at the photo. please explain answer

PLEASE HELP I WILL GIVE BRAINLIEST Use the rules of equations and inverse operations to solve the equation.

Leno4ka [110]

7 because I just know like how kould you not know that

6 0

3 years ago

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

4 years ago

A

gavmur [86]

Answer:

m∠QPM=43°

Step-by-step explanation:

see the attached figure to better understand the problem

we know that

m∠NPQ=m∠MPN+m∠MPQ

we have

m∠NPQ=(9x-25)°

m∠MPN=(4x+12)°

m∠MPQ=(3x-5)°

substitute the given values and solve for x

(9x-25)°=(4x+12)°+(3x-5)°

(9x-25)°=(7x+7)°

9x-7x=25+7

2x=32

x=16

Find the measure of angle QPM

Remember that

m∠QPM=m∠MPQ

m∠MPQ=(3x-5)°

substitute the value of x

m∠MPQ=(3(16)-5)=43°

therefore

m∠QPM=43°

7 0

4 years ago

Question 1: Four expressions are shown below:

vekshin1

Answers;

Question 1 answer: The first and last option.

Question 1 explanation: 2(4x + 2) is 4x = 12 = 2 = 14 x 2 = 28 and 8x + 4 is 8 x 4 = 32 + 4 = 36 which is 8 more than 28 then, 3x = 9 + 2 + 3 x 2 = 28.

2(4x + 2) = 28 and 8x + 4 equals 36 which is 8 more and they're both equivalent to 2(3x + 2 + x) because 2(3x = 2 = x) equals 28.

6 0

3 years ago

Fiona finds that the distance between two cities is 83,400,000 millimeters. What is the distance between the two cities in kilom

daser333 [38]

The metric table goes millimeters, centimeters, decimeters, meters, dekameters, hectometers, kilometers. To move up the table, move the decimal point one to the left. If you wanted to change it to centimeters, the answer would be 8,340,000. In kilometers, this is 83.4 kilometers. Hope this helps!

8 0

3 years ago

Read 2 more answers