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3 years ago

I need to find the area. please help

Mathematics

2 answers:

Galina-37 [17]3 years ago

8 0

Answer:

A = 190 inches^2

Step-by-step explanation:

The total area would be area for a rectangle + <em>half</em> the area of a circle:

Area for a rectangle:

A = l x w

A = 12 x 15

A = 180 $inches^{2}$

Area for half of the circle:

A = $\frac{\pi r^{2} }{2}$

$A = \frac{\pi 2.5^{2} }{2} \\\\= \frac{19.63}{2} \\\\= 9.8 inches^{2}$

Total area: Rectangle area + half of the circle area

= 180 + 9.8 = 189.8 = 190 inches^2

rusak2 [61]3 years ago

3 0

Step-by-step explanation:

Find the rectangle area first by using the formula A=length * width then find the semicircle area by using the formula A=pi*r²/2

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3 years ago

The present age of the father is twice the age of his daughter. Ten years ago, the

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Answer:

<h3> The father is 40 and the daughter is 20.</h3>

Step-by-step explanation:

x - the present age of the daughter

2x - the present age of the fathter

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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 <em>X</em> 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 <em>X</em> = number of games played.

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

(a)

The random variable <em>X</em> follows a Geometric distribution.

The probability mass function of <em>X</em> 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 <em>X</em> 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.

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