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3 years ago

Whats the definition of right triangle

Mathematics

2 answers:

OlgaM077 [116]3 years ago

5 0

Answer:

a triangle with a right angle

Step-by-step explanation:

LuckyWell [14K]3 years ago

5 0

Answer: A triangle in which one of the interior angles is 90° is called a right triangle. The longest side of the right triangle, which is also the side opposite the right angle, is the hypotenuse and the two arms of the right angle are the height and the base.

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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.

(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

Log(2)4+log(2)3,16227766/log(2)20+log(2)8

Shalnov [3]

We know the rules of logarithms:
log a^b = b * log a
log a * b = log a + log b
and we know that : 10^0.5 = 3,166227766
So:
( log (2) 4 + log (2) 3,16227766) / ( log (2) 20 + log (2) 8 ) =
= ( 2 + log (2) 10^(0.5) ) / ( log (2) 2 + log (2) 10 + 3 ) =
= ( 2 + 0.5 log (2) 10 ) / ( 1 + log (2) 10 + 3 ) =
= ( 2 + 0.5 log (2) 10 / ( 4 + log (2) 10 ) =
= ( 2 + 0.5 log (2) 10 ) / 2 * ( 2 + 0.5 log (2) 10 ) = ( cancelling )
= 1 / 2 = 0.5

3 0

3 years ago

 (6 pts) The average age of CEOs is 56 years. Assume the variable is normally distributed. If the SD is four years, find the pr

Alenkinab [10]

Answer:

The probability that the age of a randomly selected CEO will be between 50 and 55 years old is 0.334.

Step-by-step explanation:

We have a normal distribution with mean=56 years and s.d.=4 years.

We have to calculate the probability that a randomly selected CEO have an age between 50 and 55.

We have to calculate the z-value for 50 and 55.

For x=50:

$z=\frac{x-\mu}{\sigma}=\frac{50-56}{4}=\frac{-6}{4}= -1.5$

For x=55:

$z=\frac{x-\mu}{\sigma}=\frac{55-56}{4}=\frac{-1}{4}=-0.25$

The probability of being between 50 and 55 years is equal to the difference between the probability of being under 55 years and the probability of being under 50 years:

$P(50$