2. A small submarine is at an elevation of -30 feet compared to sea level. What is its elevation att

Find the volume of the triangular prism pictured below with a=10 cm , b=9 cm , and h=8 cm

erma4kov [3.2K]

Answer:

V=720 cm^3

Step-by-step explanation:

8 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

3 years ago

Find the area. 4 5 square units

Ymorist [56]

Answer: 20 Square units

Explanation: A= b x h = 4 x 5 = 20

3 0

3 years ago

The combined age of three relatives is 120 years. James is three times the age of Dan, and Paul is two times the sum of the ages

d1i1m1o1n [39]

Answer:

Step-by-step explanation:

From the problem statement, we can create the following equation:

$J + P + D = 120$

where $J$ is the age of James, $P$ is the age of Paul, and $D$ is the age of Dan.

From the first part of the second sentence, we can set up the following equation:

$J = 3D$

From the last part of the second sentence, we can set up the following equation:

$P = 2(J + D)$

We can substitute the second equation into the last one to get the following:

$P = 2(3D + D)$

$P = 2(4D)$

$P = 8D$

We can then substitute the last two equations in the first to solve for $D$ :

$3D + 8D + D = 120$

$12D = 120$

$D = 10$

Plugging this into the other two equations will give us the remaining ages:

$P = 8D$

$P = 8(10)$

$P = 80$

$J = 3D$

$J = 3(10)$

$J = 30$

8 0

3 years ago

Does anyone know this??

AleksAgata [21]

Answer:

slopes

Step-by-step explanation:

6 0

3 years ago