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3 years ago

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Two containers designed to hold water are side by side, both in the shape of a cylinder. Container A has a diameter of 8 feet an

d a height of 8 feet. Container B has a diameter of 6 feet and a height of 13 feet. Container A is full of water and the water is pumped into Container B until Conainter B is completely full. To the nearest tenth, what is the percent of Container A that is empty after the pumping is complete?

1 answer:

9966 [12]3 years ago

4 0

Answer: 91.4%

Step-by-step explanation:

The formula for determining the volume of a cylinder is expressed as

Volume = πr²h

Where

r represents the radius of the cylinder.

h represents the height of the cylinder.

π is a constant whose value is 3.14

Considering cylinder A,

Height = 8ft

Diameter = 8 ft

Radius = diameter/2 = 8/2 = 4ft

Therefore,

Volume = 3.14 × 4² × 8 = 401.92ft³

Considering cylinder B,

Height = 13ft

Diameter = 6 ft

Radius = diameter/2 = 6/2 = 3ft

Therefore,

Volume = 3.14 × 3² × 13 = 367.38ft³

The volume of Container A that would be left empty after container B is pumped into it is

367.38ft³

the percent of Container A that is empty after the pumping is complete is

367.38/401.92 × 100 = 91.4%

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Step-by-step explanation:

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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.

(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 <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.

4 0

3 years ago

1. What is the value of the expression 3^-4?. a) -81. b) -12. c) -1/81. d) 1/81. 2. What is the value of the expression 1/4^3?.

hjlf

Answer:

1.

Option d is correct

2.

Option c is correct

3.

Option a is correct

4.

Option d is correct

5.

Option c is correct

Explanation:

Using exponent rule:

$a^{-n} = \frac{1}{a^n}$

1.

Given the expression:

$3^{-4}$

Apply the exponent rules:

$\frac{1}{3^4} = \frac{1}{81}$

Therefore, the value of the given expression is, $\frac{1}{81}$ .

2.

To find the value of the expression $\frac{1}{4^3}$

then;

$\frac{1}{4^3} = \frac{1}{64}$

Therefore, the value of the given expression is, $\frac{1}{64}$ .

3.

Find the the value of the expression $\frac{2}{4^3}$

then;

$\frac{2}{4^3} = \frac{2}{64}$

Simplify:

$\frac{1}{32}$

Therefore, the value of the given expression is, $\frac{1}{32}$ .

4.

Given the expression:

$K^{-3} = \frac{1}{27}$

we can write this as:

$K^{-3} = \frac{1}{3^3}$

Apply the exponent rules:

$K^{-3} = 3^{-3}$

On comparing both sides we have;

K = 3

Therefore, the value of K is, 3

5.

Given the expression:

$3^N= \frac{1}{81}$

we can write this as:

$3^N= \frac{1}{3^4}$

Apply the exponent rules:

$3^N = 3^{-4}$

On comparing both sides we have;

N = 4

Therefore, the value of N is, -4

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4 years ago

Read 2 more answers

A recipe calls for 2 tsp of salt, 1 tsp of pepper, and 4 tsp of garlic powder. How much pepper and garlic powder would you need

MakcuM [25]

Answer:

amount of pepper required= 7.5 tsp

amount of garlic powder required = 30 tsp

Step-by-step explanation:

Given,

amount of salt used for small batch of the recipe = 2 tsp

amount of pepper used for small batch of the recipe = 1 tsp

amount of garlic powder used for small batch of the recipe = 4 tsp

amount of salt used for the larger batch = 15 tsp

= 2 x 7.5 tsp

= amount of salt used for small batch the recipe x 7.5

So,

the amount of pepper needed for the larger batch= 7.5 x amount of pepper used for the small batch of recipe

= 7.5 x 1 tsp

= 7.5 tsp

the amount of garlic powder needed for the larger batch= 7.5 x amount of garlic powder used for the small batch of recipe

= 7.5 x 4 tsp

= 30 tsp

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3 years ago

A tank currently holds 5 gallons of water, it being filled at the rate of 1/4 gallon per minute. How long will it take for the t

Blababa [14]

Answer:

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Step-by-step explanation:

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4 years ago